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Suppose that a teacher wishes to distribute 25 identical pencils to Ahmed, Barbara, Casper, and Dieter such that Ahmed and Dieter receive at least one pencil each, Casper receives no more than five pencils, and Barbara receives at least four pencils. In how many ways can such a distribution be made?

User Bmatovu
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Answer:

we will change the problem in stages

pre-place 1 in x1 & x4, and 4 in x2, thus guaranteeing their minimum requirements.

now 19 pencils remain to be distributed

when there are no restrictions on distributing n identical objects to k distinct piles, the formula is (n+k-1)C(k-1) which here translates to 22C3

finally, we need to see that x3 = 5, so we place 6 in x3 (guaranteeing that it violates the condition), and subtract such violations, ie (22-6)C3 = 16C3

ans: 22C3 - 16C3 = 980

User Sam Parrish
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