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How many ways can you distribute $10$ identical balls into $5$ distinguishable boxes so that the sum total of balls in the first two boxes equals $6$?

This is probability/Combinatorics

What I have so far:

$$\frac{6!}{2!(6-2)!} + \frac{4!}{3!(4-3)!} = 19$$

Does this look correct?

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  • $\begingroup$ Formatting tips here. $\endgroup$ – Em. Feb 15 '16 at 3:33
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*Hint:

For problems of identical balls into distinguishable boxes, you must be familiar with stars and bars:

The problem needs to be broken into $2$ parts: $6$ balls in the first two boxes, $4$ balls in the next three, apply stars and bars for each, and then apply the multiplication principle.

For the first part, taking zero balls in boxes to be permissible, the number of ways = $\binom{6+2-1}{2-1} = 7$, you can verify the formula for this simple case by enumeration.

Proceed.....

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    $\begingroup$ $\binom{6+2-1}{2-1} = 7$ and $\binom{4+3-1}{3-1} = 15$. Multiplying both I get 105. Is this on the right track? $\endgroup$ – Dangerous Game Feb 15 '16 at 4:19
  • $\begingroup$ Yes. You should develop confidence to proceed, given sufficient hint. $\endgroup$ – true blue anil Feb 15 '16 at 4:24
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You have two disjoint problems: divide six balls between the first two boxes (how many ways?) and divide the other four between the remaining three boxes. As they are independent, you multiply the results for the two problems. As $19$ is prime, it is unlikely to be the correct answer.

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  • $\begingroup$ Should it be just (6C2) * (4C3) = 60? $\endgroup$ – Dangerous Game Feb 15 '16 at 4:08
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    $\begingroup$ No, because the balls are identical. $6C2=15$ claims there are that many ways to divide the six balls into two bins, but you can only pick $0$ to $6$ to be in the first bin, then the second is determined for $7$. To get the second set, you need stars and bars as true blue anil says. $\endgroup$ – Ross Millikan Feb 15 '16 at 4:11
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The number of ways to distribute 10 identical balls into 5 distinguishable boxes is often solved using the “stars and bars” technique. Ten stars represent the balls, and four bars represent the dividing points between the first and second, second and third, third and fourth, and fourth and fifth boxes.

Thus the arrangement of 10 *s and 4 bars **|*****||*|** represents putting 2 balls in the first box, 5 in the second box, none in the third box, one in the fourth box, and two in the fifth box.

An arrangement where the total of balls in the first two boxes is 6 must have six *s and one | before the second | (which separates the first two boxes from the last three). The second | is therefore the 8th symbol in the arrangement.

So you need to count the arrangements of 10 *s and 4 bars that have the pattern [any arrangement of six *s and one 1] | [any arrangement of four *s and two bars].

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