If $C(16,r) = C(16,r+2)$, then find $r$. Explain how you know.












2












$begingroup$



If $C(16,r) = C(16,r+2)$, then find $r$. Explain how you know.




Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?






combinations







share|cite|improve this question











$endgroup$

















    2












    $begingroup$



    If $C(16,r) = C(16,r+2)$, then find $r$. Explain how you know.




    Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?






    combinations







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$



      If $C(16,r) = C(16,r+2)$, then find $r$. Explain how you know.




      Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?






      combinations







      share|cite|improve this question











      $endgroup$





      If $C(16,r) = C(16,r+2)$, then find $r$. Explain how you know.




      Just started dealing with combinations and permutations rights now and came across this study problem. I've converted each side into their combination formula forms, but I'm not sure where to go from there(or even if that's what I should be doing). Any help with this or at least where I should start?






      combinations




      combinations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 9:23









      Tianlalu

      3,08621038




      3,08621038










      asked Dec 3 '18 at 4:23









      Gold Pony BoyGold Pony Boy

      152




      152






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          Note that $binom{n}{k} = binom{n}{n-k}.$



          What equation does this give for $r$ in your scenario? What is the solution to that equation?



          Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Your equation does not imply $$binom na=binom nbimplies a=n-b$$
            $endgroup$
            – Kemono Chen
            Dec 3 '18 at 4:52












          • $begingroup$
            @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
            $endgroup$
            – Display name
            Dec 3 '18 at 5:00












          • $begingroup$
            Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
            $endgroup$
            – Kemono Chen
            Dec 3 '18 at 5:01



















          4












          $begingroup$

          Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Okay, I see the connection now, so r must be 7...?
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 4:45










          • $begingroup$
            @GoldPonyBoy Yes.
            $endgroup$
            – Carl Schildkraut
            Dec 3 '18 at 6:06



















          2












          $begingroup$

          $C_r^{16}=C_{r+2}^{16}$
          Then using the definition of combination you'll get



          $$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
          $$implies frac{1}{r!(16-r)(15-r).(14-r)!}=frac{1}{(r+2)(r+1).r!.(14-r)!}$$
          $$implies (16-r)(15-r)=(r+2)(r+1)$$
          $$implies 240-31r+r^2=r^2+3r+2$$
          $$implies 34r=238$$
          $$implies r=7$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 5:06






          • 2




            $begingroup$
            @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
            $endgroup$
            – mathematics2x2life
            Dec 3 '18 at 5:07










          • $begingroup$
            @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 5:12








          • 1




            $begingroup$
            @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
            $endgroup$
            – Fareed AF
            Dec 3 '18 at 5:30






          • 1




            $begingroup$
            This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
            $endgroup$
            – smci
            Dec 3 '18 at 17:46













          Your Answer





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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Note that $binom{n}{k} = binom{n}{n-k}.$



          What equation does this give for $r$ in your scenario? What is the solution to that equation?



          Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Your equation does not imply $$binom na=binom nbimplies a=n-b$$
            $endgroup$
            – Kemono Chen
            Dec 3 '18 at 4:52












          • $begingroup$
            @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
            $endgroup$
            – Display name
            Dec 3 '18 at 5:00












          • $begingroup$
            Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
            $endgroup$
            – Kemono Chen
            Dec 3 '18 at 5:01
















          3












          $begingroup$

          Note that $binom{n}{k} = binom{n}{n-k}.$



          What equation does this give for $r$ in your scenario? What is the solution to that equation?



          Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Your equation does not imply $$binom na=binom nbimplies a=n-b$$
            $endgroup$
            – Kemono Chen
            Dec 3 '18 at 4:52












          • $begingroup$
            @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
            $endgroup$
            – Display name
            Dec 3 '18 at 5:00












          • $begingroup$
            Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
            $endgroup$
            – Kemono Chen
            Dec 3 '18 at 5:01














          3












          3








          3





          $begingroup$

          Note that $binom{n}{k} = binom{n}{n-k}.$



          What equation does this give for $r$ in your scenario? What is the solution to that equation?



          Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.






          share|cite|improve this answer











          $endgroup$



          Note that $binom{n}{k} = binom{n}{n-k}.$



          What equation does this give for $r$ in your scenario? What is the solution to that equation?



          Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$ Since $r ne r+2,$ the value of $r$ that you find is unique.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 '18 at 6:35

























          answered Dec 3 '18 at 4:28









          Display nameDisplay name

          836314




          836314












          • $begingroup$
            Your equation does not imply $$binom na=binom nbimplies a=n-b$$
            $endgroup$
            – Kemono Chen
            Dec 3 '18 at 4:52












          • $begingroup$
            @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
            $endgroup$
            – Display name
            Dec 3 '18 at 5:00












          • $begingroup$
            Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
            $endgroup$
            – Kemono Chen
            Dec 3 '18 at 5:01


















          • $begingroup$
            Your equation does not imply $$binom na=binom nbimplies a=n-b$$
            $endgroup$
            – Kemono Chen
            Dec 3 '18 at 4:52












          • $begingroup$
            @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
            $endgroup$
            – Display name
            Dec 3 '18 at 5:00












          • $begingroup$
            Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
            $endgroup$
            – Kemono Chen
            Dec 3 '18 at 5:01
















          $begingroup$
          Your equation does not imply $$binom na=binom nbimplies a=n-b$$
          $endgroup$
          – Kemono Chen
          Dec 3 '18 at 4:52






          $begingroup$
          Your equation does not imply $$binom na=binom nbimplies a=n-b$$
          $endgroup$
          – Kemono Chen
          Dec 3 '18 at 4:52














          $begingroup$
          @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
          $endgroup$
          – Display name
          Dec 3 '18 at 5:00






          $begingroup$
          @KemonoChen Note that $binom{n}{a+1} > binom{n}{a}$ when $a < lfloor n/2 rfloor,$ hence an integer can be taken on at most twice as you range among $binom{n}{0}, dots, binom{n}{n}.$
          $endgroup$
          – Display name
          Dec 3 '18 at 5:00














          $begingroup$
          Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
          $endgroup$
          – Kemono Chen
          Dec 3 '18 at 5:01




          $begingroup$
          Exactly. I think it would be better to put this into your answer to clarify. :P The original answer is logically incorrect.
          $endgroup$
          – Kemono Chen
          Dec 3 '18 at 5:01











          4












          $begingroup$

          Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Okay, I see the connection now, so r must be 7...?
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 4:45










          • $begingroup$
            @GoldPonyBoy Yes.
            $endgroup$
            – Carl Schildkraut
            Dec 3 '18 at 6:06
















          4












          $begingroup$

          Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Okay, I see the connection now, so r must be 7...?
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 4:45










          • $begingroup$
            @GoldPonyBoy Yes.
            $endgroup$
            – Carl Schildkraut
            Dec 3 '18 at 6:06














          4












          4








          4





          $begingroup$

          Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?






          share|cite|improve this answer









          $endgroup$



          Hint: Think about row $16$ of Pascal's triangle. When does it increase (going to the right) and when does it decrease? What values are duplicated and where?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 3 '18 at 4:29









          Carl SchildkrautCarl Schildkraut

          11.2k11441




          11.2k11441












          • $begingroup$
            Okay, I see the connection now, so r must be 7...?
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 4:45










          • $begingroup$
            @GoldPonyBoy Yes.
            $endgroup$
            – Carl Schildkraut
            Dec 3 '18 at 6:06


















          • $begingroup$
            Okay, I see the connection now, so r must be 7...?
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 4:45










          • $begingroup$
            @GoldPonyBoy Yes.
            $endgroup$
            – Carl Schildkraut
            Dec 3 '18 at 6:06
















          $begingroup$
          Okay, I see the connection now, so r must be 7...?
          $endgroup$
          – Gold Pony Boy
          Dec 3 '18 at 4:45




          $begingroup$
          Okay, I see the connection now, so r must be 7...?
          $endgroup$
          – Gold Pony Boy
          Dec 3 '18 at 4:45












          $begingroup$
          @GoldPonyBoy Yes.
          $endgroup$
          – Carl Schildkraut
          Dec 3 '18 at 6:06




          $begingroup$
          @GoldPonyBoy Yes.
          $endgroup$
          – Carl Schildkraut
          Dec 3 '18 at 6:06











          2












          $begingroup$

          $C_r^{16}=C_{r+2}^{16}$
          Then using the definition of combination you'll get



          $$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
          $$implies frac{1}{r!(16-r)(15-r).(14-r)!}=frac{1}{(r+2)(r+1).r!.(14-r)!}$$
          $$implies (16-r)(15-r)=(r+2)(r+1)$$
          $$implies 240-31r+r^2=r^2+3r+2$$
          $$implies 34r=238$$
          $$implies r=7$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 5:06






          • 2




            $begingroup$
            @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
            $endgroup$
            – mathematics2x2life
            Dec 3 '18 at 5:07










          • $begingroup$
            @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 5:12








          • 1




            $begingroup$
            @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
            $endgroup$
            – Fareed AF
            Dec 3 '18 at 5:30






          • 1




            $begingroup$
            This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
            $endgroup$
            – smci
            Dec 3 '18 at 17:46


















          2












          $begingroup$

          $C_r^{16}=C_{r+2}^{16}$
          Then using the definition of combination you'll get



          $$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
          $$implies frac{1}{r!(16-r)(15-r).(14-r)!}=frac{1}{(r+2)(r+1).r!.(14-r)!}$$
          $$implies (16-r)(15-r)=(r+2)(r+1)$$
          $$implies 240-31r+r^2=r^2+3r+2$$
          $$implies 34r=238$$
          $$implies r=7$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 5:06






          • 2




            $begingroup$
            @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
            $endgroup$
            – mathematics2x2life
            Dec 3 '18 at 5:07










          • $begingroup$
            @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 5:12








          • 1




            $begingroup$
            @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
            $endgroup$
            – Fareed AF
            Dec 3 '18 at 5:30






          • 1




            $begingroup$
            This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
            $endgroup$
            – smci
            Dec 3 '18 at 17:46
















          2












          2








          2





          $begingroup$

          $C_r^{16}=C_{r+2}^{16}$
          Then using the definition of combination you'll get



          $$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
          $$implies frac{1}{r!(16-r)(15-r).(14-r)!}=frac{1}{(r+2)(r+1).r!.(14-r)!}$$
          $$implies (16-r)(15-r)=(r+2)(r+1)$$
          $$implies 240-31r+r^2=r^2+3r+2$$
          $$implies 34r=238$$
          $$implies r=7$$






          share|cite|improve this answer











          $endgroup$



          $C_r^{16}=C_{r+2}^{16}$
          Then using the definition of combination you'll get



          $$frac{16!}{r!(16-r)!}=frac{16!}{(r+2)!(16-r-2)!}$$
          $$implies frac{1}{r!(16-r)(15-r).(14-r)!}=frac{1}{(r+2)(r+1).r!.(14-r)!}$$
          $$implies (16-r)(15-r)=(r+2)(r+1)$$
          $$implies 240-31r+r^2=r^2+3r+2$$
          $$implies 34r=238$$
          $$implies r=7$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 '18 at 6:50

























          answered Dec 3 '18 at 4:41









          Fareed AFFareed AF

          45211




          45211












          • $begingroup$
            Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 5:06






          • 2




            $begingroup$
            @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
            $endgroup$
            – mathematics2x2life
            Dec 3 '18 at 5:07










          • $begingroup$
            @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 5:12








          • 1




            $begingroup$
            @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
            $endgroup$
            – Fareed AF
            Dec 3 '18 at 5:30






          • 1




            $begingroup$
            This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
            $endgroup$
            – smci
            Dec 3 '18 at 17:46




















          • $begingroup$
            Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 5:06






          • 2




            $begingroup$
            @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
            $endgroup$
            – mathematics2x2life
            Dec 3 '18 at 5:07










          • $begingroup$
            @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
            $endgroup$
            – Gold Pony Boy
            Dec 3 '18 at 5:12








          • 1




            $begingroup$
            @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
            $endgroup$
            – Fareed AF
            Dec 3 '18 at 5:30






          • 1




            $begingroup$
            This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
            $endgroup$
            – smci
            Dec 3 '18 at 17:46


















          $begingroup$
          Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
          $endgroup$
          – Gold Pony Boy
          Dec 3 '18 at 5:06




          $begingroup$
          Sorry, I must be stupid or something and my algebra is off, but i can't get from the last line to r = 7
          $endgroup$
          – Gold Pony Boy
          Dec 3 '18 at 5:06




          2




          2




          $begingroup$
          @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
          $endgroup$
          – mathematics2x2life
          Dec 3 '18 at 5:07




          $begingroup$
          @GoldPonyBoy Just expand both sides, move everything to one side, and solve the quadratic equation.
          $endgroup$
          – mathematics2x2life
          Dec 3 '18 at 5:07












          $begingroup$
          @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
          $endgroup$
          – Gold Pony Boy
          Dec 3 '18 at 5:12






          $begingroup$
          @mathematics2x2life Oh I AM just dumb, I wrote -r*-r = -r^2. This is what I started to do originally, but couldn't follow the logic at first. Thanks for the help, I really appreciate it.
          $endgroup$
          – Gold Pony Boy
          Dec 3 '18 at 5:12






          1




          1




          $begingroup$
          @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
          $endgroup$
          – Fareed AF
          Dec 3 '18 at 5:30




          $begingroup$
          @Gold Pony Boy we have that $(r+2)!=(r+2)(r+1).r!$ So you remove $r!$ both sides and you are left with $(r+1)(r+2)$ on the right side
          $endgroup$
          – Fareed AF
          Dec 3 '18 at 5:30




          1




          1




          $begingroup$
          This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
          $endgroup$
          – smci
          Dec 3 '18 at 17:46






          $begingroup$
          This approach is true but massive overkill. Better to note that the denominator $D(r) = r!(16−r)!$ is monotonic increasing for r < floor(n/2), achieves a maximum at n/2 = 8, and is symmetric about floor(n/2). From that we can directly conclude that if D(x) = D(y), then we must have y = n-x
          $endgroup$
          – smci
          Dec 3 '18 at 17:46




















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