To find the number of ways to put 14 identical balls into 4 bins with the condition that no bin can

Kellen Perkins

Kellen Perkins

Answered question

2022-05-20

To find the number of ways to put 14 identical balls into 4 bins with the condition that no bin can hold more than 7 balls.
I have tried the following:
The total no of ways to distribute 14 identical balls into 4 bins without any restriction is
( 14 + 4 1 4 1 ) = ( 17 3 ) ..
Note that there can't be two bins with more than 7 balls since we have only 14 identical balls only.
Now, we count the no. of ways so that one bin has more than 7 balls. So, it has at least 8 balls and the remaining 6 can be distributed in ( 6 + 4 1 4 1 ) = ( 9 3 ) ways. We can choose one bin out of 4 in 4 ways.
Hence the reqd number of ways = ( 17 3 ) 4 × ( 9 3 ) .

Answer & Explanation

Makai Blackwell

Makai Blackwell

Beginner2022-05-21Added 11 answers

Step 1
Your required sum is the coefficient of x 14 in
( 1 + x + + x 7 ) 4
or [ x 14 ] : ( 1 x 8 1 x ) 4
Step 2
We can expand ( 1 x 8 ) 4 into 1 4 x 8 , and the other term is i = 0 ( i + 3 3 ) x i .
We want the values only when i = 6 or i = 14, which gives your equation.
Trevor Wood

Trevor Wood

Beginner2022-05-22Added 5 answers

Step 1
What you have is correct. Here’s a slightly alternative approach. You want to count the nonnegative integer solutions to x 1 + x 2 + x 3 + x 4 = 14 such that x i 7 for all i. By inclusion-exclusion and stars-and-bars, we have
( 14 + 4 1 4 1 ) ( 4 1 ) k = 8 14 ( 14 k + 3 1 3 1 )
solutions. By the hockey-stick identity, the sum reduces to ( 9 3 ) as in your answer.

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