How to distribute 5 cards among 3 people $A,B,C$ so that none of the person have empty hands
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How to distribute 5 cards among 3 people $A,B,C$ so that none of the person have empty hands?
My try:
5 cards can be distributed among $A,B,C$ in $2,2,1$ ways.
Now there are three ways of arranging $2,2,1$ .
Once I decide whether A or B or C is 2 or 1 we have $binom 5 2$ ways of distributing 2 cards to the person who has 2 cards ,$binom 3 2$ ways of distributing 2 cards to another person who has 2 cards
Thus the answer is $binom 5 2times binom 3 2times 3$
Is the answer correct??
Please correct if not .
combinatorics number-theory divisor-counting-function
$endgroup$
add a comment |
$begingroup$
How to distribute 5 cards among 3 people $A,B,C$ so that none of the person have empty hands?
My try:
5 cards can be distributed among $A,B,C$ in $2,2,1$ ways.
Now there are three ways of arranging $2,2,1$ .
Once I decide whether A or B or C is 2 or 1 we have $binom 5 2$ ways of distributing 2 cards to the person who has 2 cards ,$binom 3 2$ ways of distributing 2 cards to another person who has 2 cards
Thus the answer is $binom 5 2times binom 3 2times 3$
Is the answer correct??
Please correct if not .
combinatorics number-theory divisor-counting-function
$endgroup$
2
$begingroup$
That's correct, but you are forgetting the case $1,1,3$
$endgroup$
– Shubham Johri
Dec 17 '18 at 15:45
add a comment |
$begingroup$
How to distribute 5 cards among 3 people $A,B,C$ so that none of the person have empty hands?
My try:
5 cards can be distributed among $A,B,C$ in $2,2,1$ ways.
Now there are three ways of arranging $2,2,1$ .
Once I decide whether A or B or C is 2 or 1 we have $binom 5 2$ ways of distributing 2 cards to the person who has 2 cards ,$binom 3 2$ ways of distributing 2 cards to another person who has 2 cards
Thus the answer is $binom 5 2times binom 3 2times 3$
Is the answer correct??
Please correct if not .
combinatorics number-theory divisor-counting-function
$endgroup$
How to distribute 5 cards among 3 people $A,B,C$ so that none of the person have empty hands?
My try:
5 cards can be distributed among $A,B,C$ in $2,2,1$ ways.
Now there are three ways of arranging $2,2,1$ .
Once I decide whether A or B or C is 2 or 1 we have $binom 5 2$ ways of distributing 2 cards to the person who has 2 cards ,$binom 3 2$ ways of distributing 2 cards to another person who has 2 cards
Thus the answer is $binom 5 2times binom 3 2times 3$
Is the answer correct??
Please correct if not .
combinatorics number-theory divisor-counting-function
combinatorics number-theory divisor-counting-function
asked Dec 17 '18 at 15:39
user596656
2
$begingroup$
That's correct, but you are forgetting the case $1,1,3$
$endgroup$
– Shubham Johri
Dec 17 '18 at 15:45
add a comment |
2
$begingroup$
That's correct, but you are forgetting the case $1,1,3$
$endgroup$
– Shubham Johri
Dec 17 '18 at 15:45
2
2
$begingroup$
That's correct, but you are forgetting the case $1,1,3$
$endgroup$
– Shubham Johri
Dec 17 '18 at 15:45
$begingroup$
That's correct, but you are forgetting the case $1,1,3$
$endgroup$
– Shubham Johri
Dec 17 '18 at 15:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
That's correct, but you are forgetting the case $1,1,3$. We can permute $1,1,3$ in $3$ ways, select 3 cards in $binom53$ ways, the next 2 cards in $binom21,binom11$ ways respectively. The total ways are $3binom53binom21binom11=60$. Add to it $3binom52binom32=90$ to get a total of $150$.
answered Dec 17 '18 at 15:50
Shubham JohriShubham Johri
5,214718
$endgroup$
$begingroup$
Oh yes ! Thank you very much
$endgroup$
– user596656
Dec 17 '18 at 16:04
add a comment |
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1 Answer
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1 Answer
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$begingroup$
That's correct, but you are forgetting the case $1,1,3$. We can permute $1,1,3$ in $3$ ways, select 3 cards in $binom53$ ways, the next 2 cards in $binom21,binom11$ ways respectively. The total ways are $3binom53binom21binom11=60$. Add to it $3binom52binom32=90$ to get a total of $150$.
answered Dec 17 '18 at 15:50
Shubham JohriShubham Johri
5,214718
$endgroup$
$begingroup$
Oh yes ! Thank you very much
$endgroup$
– user596656
Dec 17 '18 at 16:04
add a comment |
$begingroup$
That's correct, but you are forgetting the case $1,1,3$. We can permute $1,1,3$ in $3$ ways, select 3 cards in $binom53$ ways, the next 2 cards in $binom21,binom11$ ways respectively. The total ways are $3binom53binom21binom11=60$. Add to it $3binom52binom32=90$ to get a total of $150$.
answered Dec 17 '18 at 15:50
Shubham JohriShubham Johri
5,214718
$endgroup$
$begingroup$
Oh yes ! Thank you very much
$endgroup$
– user596656
Dec 17 '18 at 16:04
add a comment |
$begingroup$
That's correct, but you are forgetting the case $1,1,3$. We can permute $1,1,3$ in $3$ ways, select 3 cards in $binom53$ ways, the next 2 cards in $binom21,binom11$ ways respectively. The total ways are $3binom53binom21binom11=60$. Add to it $3binom52binom32=90$ to get a total of $150$.
answered Dec 17 '18 at 15:50
Shubham JohriShubham Johri
5,214718
$endgroup$
That's correct, but you are forgetting the case $1,1,3$. We can permute $1,1,3$ in $3$ ways, select 3 cards in $binom53$ ways, the next 2 cards in $binom21,binom11$ ways respectively. The total ways are $3binom53binom21binom11=60$. Add to it $3binom52binom32=90$ to get a total of $150$.
answered Dec 17 '18 at 15:50
Shubham JohriShubham Johri
5,214718
answered Dec 17 '18 at 15:50
Shubham JohriShubham Johri
5,214718
answered Dec 17 '18 at 15:50
Shubham JohriShubham Johri
5,214718
answered Dec 17 '18 at 15:50
Shubham JohriShubham Johri
5,214718
5,214718
$begingroup$
Oh yes ! Thank you very much
$endgroup$
– user596656
Dec 17 '18 at 16:04
add a comment |
$begingroup$
Oh yes ! Thank you very much
$endgroup$
– user596656
Dec 17 '18 at 16:04
$begingroup$
Oh yes ! Thank you very much
$endgroup$
– user596656
Dec 17 '18 at 16:04
$begingroup$
Oh yes ! Thank you very much
$endgroup$
– user596656
Dec 17 '18 at 16:04
add a comment |
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$begingroup$
That's correct, but you are forgetting the case $1,1,3$
$endgroup$
– Shubham Johri
Dec 17 '18 at 15:45