Confusing combination-permutations question
$begingroup$
In a shop there are five types of ice-creams available. A child buys
six ice-creams. Is it true that the number of different ways the child
can buy six ice creams is equal to the number of different ways of
arranging 6 A's and 4 B's in a row?
I tried it as follows:
The child problem-
There are five kinds of ice-creams, so the child can select six ice-creams in $5^6=15625$ ways.
A's and B's problem-
There are 6 A's and 4 B's. We can place these 10 items in a row in 10!/6!4! ways, which is equal to 210.
But the actual answer is true. How can it be? Where am I wrong?
permutations
asked Apr 3 '13 at 15:44
syfluqssyfluqs
42
$endgroup$
add a comment |
$begingroup$
In a shop there are five types of ice-creams available. A child buys
six ice-creams. Is it true that the number of different ways the child
can buy six ice creams is equal to the number of different ways of
arranging 6 A's and 4 B's in a row?
I tried it as follows:
The child problem-
There are five kinds of ice-creams, so the child can select six ice-creams in $5^6=15625$ ways.
A's and B's problem-
There are 6 A's and 4 B's. We can place these 10 items in a row in 10!/6!4! ways, which is equal to 210.
But the actual answer is true. How can it be? Where am I wrong?
permutations
asked Apr 3 '13 at 15:44
syfluqssyfluqs
42
$endgroup$
add a comment |
$begingroup$
In a shop there are five types of ice-creams available. A child buys
six ice-creams. Is it true that the number of different ways the child
can buy six ice creams is equal to the number of different ways of
arranging 6 A's and 4 B's in a row?
I tried it as follows:
The child problem-
There are five kinds of ice-creams, so the child can select six ice-creams in $5^6=15625$ ways.
A's and B's problem-
There are 6 A's and 4 B's. We can place these 10 items in a row in 10!/6!4! ways, which is equal to 210.
But the actual answer is true. How can it be? Where am I wrong?
permutations
asked Apr 3 '13 at 15:44
syfluqssyfluqs
42
$endgroup$
In a shop there are five types of ice-creams available. A child buys
six ice-creams. Is it true that the number of different ways the child
can buy six ice creams is equal to the number of different ways of
arranging 6 A's and 4 B's in a row?
I tried it as follows:
The child problem-
There are five kinds of ice-creams, so the child can select six ice-creams in $5^6=15625$ ways.
A's and B's problem-
There are 6 A's and 4 B's. We can place these 10 items in a row in 10!/6!4! ways, which is equal to 210.
But the actual answer is true. How can it be? Where am I wrong?
permutations
permutations
asked Apr 3 '13 at 15:44
syfluqssyfluqs
42
asked Apr 3 '13 at 15:44
syfluqssyfluqs
42
asked Apr 3 '13 at 15:44
syfluqssyfluqs
42
asked Apr 3 '13 at 15:44
syfluqssyfluqs
42
asked Apr 3 '13 at 15:44
syfluqssyfluqs
42
42
add a comment |
add a comment |
1 Answer
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$begingroup$
The answer provided by you is wrong. The count of $5^6$ selects and arranges 6 ice-creams in a order after selecting from 5 different ice-creams. But we want to get no of ways by which the child can buy 6 ice-creams and not to arrange them.
The correct approach can be as follows as the total no of ice creams of each type are unknown .
Let no of ice-creams bought of each type as $a,b,c,d,e$ .So, $$a+b+c+d+e=6$$ as boy needs 6 ice-creams. And thus answer is $10_{C_4}$ .
Which is same as arranging 6A's and 4B's in a row.
answered Apr 3 '13 at 16:13
ABCABC
5,14531438
$endgroup$
$begingroup$
Here is my method to get no. of solutions for $a+b+c=k$ types:math.stackexchange.com/questions/338525/…
$endgroup$
– ABC
Apr 3 '13 at 16:26
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
The answer provided by you is wrong. The count of $5^6$ selects and arranges 6 ice-creams in a order after selecting from 5 different ice-creams. But we want to get no of ways by which the child can buy 6 ice-creams and not to arrange them.
The correct approach can be as follows as the total no of ice creams of each type are unknown .
Let no of ice-creams bought of each type as $a,b,c,d,e$ .So, $$a+b+c+d+e=6$$ as boy needs 6 ice-creams. And thus answer is $10_{C_4}$ .
Which is same as arranging 6A's and 4B's in a row.
answered Apr 3 '13 at 16:13
ABCABC
5,14531438
$endgroup$
$begingroup$
Here is my method to get no. of solutions for $a+b+c=k$ types:math.stackexchange.com/questions/338525/…
$endgroup$
– ABC
Apr 3 '13 at 16:26
add a comment |
$begingroup$
The answer provided by you is wrong. The count of $5^6$ selects and arranges 6 ice-creams in a order after selecting from 5 different ice-creams. But we want to get no of ways by which the child can buy 6 ice-creams and not to arrange them.
The correct approach can be as follows as the total no of ice creams of each type are unknown .
Let no of ice-creams bought of each type as $a,b,c,d,e$ .So, $$a+b+c+d+e=6$$ as boy needs 6 ice-creams. And thus answer is $10_{C_4}$ .
Which is same as arranging 6A's and 4B's in a row.
answered Apr 3 '13 at 16:13
ABCABC
5,14531438
$endgroup$
$begingroup$
Here is my method to get no. of solutions for $a+b+c=k$ types:math.stackexchange.com/questions/338525/…
$endgroup$
– ABC
Apr 3 '13 at 16:26
add a comment |
$begingroup$
The answer provided by you is wrong. The count of $5^6$ selects and arranges 6 ice-creams in a order after selecting from 5 different ice-creams. But we want to get no of ways by which the child can buy 6 ice-creams and not to arrange them.
The correct approach can be as follows as the total no of ice creams of each type are unknown .
Let no of ice-creams bought of each type as $a,b,c,d,e$ .So, $$a+b+c+d+e=6$$ as boy needs 6 ice-creams. And thus answer is $10_{C_4}$ .
Which is same as arranging 6A's and 4B's in a row.
answered Apr 3 '13 at 16:13
ABCABC
5,14531438
$endgroup$
The answer provided by you is wrong. The count of $5^6$ selects and arranges 6 ice-creams in a order after selecting from 5 different ice-creams. But we want to get no of ways by which the child can buy 6 ice-creams and not to arrange them.
The correct approach can be as follows as the total no of ice creams of each type are unknown .
Let no of ice-creams bought of each type as $a,b,c,d,e$ .So, $$a+b+c+d+e=6$$ as boy needs 6 ice-creams. And thus answer is $10_{C_4}$ .
Which is same as arranging 6A's and 4B's in a row.
answered Apr 3 '13 at 16:13
ABCABC
5,14531438
edited Apr 3 '13 at 16:20
answered Apr 3 '13 at 16:13
ABCABC
5,14531438
answered Apr 3 '13 at 16:13
ABCABC
5,14531438
answered Apr 3 '13 at 16:13
ABCABC
5,14531438
5,14531438
$begingroup$
Here is my method to get no. of solutions for $a+b+c=k$ types:math.stackexchange.com/questions/338525/…
$endgroup$
– ABC
Apr 3 '13 at 16:26
add a comment |
$begingroup$
Here is my method to get no. of solutions for $a+b+c=k$ types:math.stackexchange.com/questions/338525/…
$endgroup$
– ABC
Apr 3 '13 at 16:26
$begingroup$
Here is my method to get no. of solutions for $a+b+c=k$ types:math.stackexchange.com/questions/338525/…
$endgroup$
– ABC
Apr 3 '13 at 16:26
$begingroup$
Here is my method to get no. of solutions for $a+b+c=k$ types:math.stackexchange.com/questions/338525/…
$endgroup$
– ABC
Apr 3 '13 at 16:26
add a comment |
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