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How many ways can I distribute 100 sweets to a class of 30 boys and 20 girls, if it is required that each boy has at least one sweet and each girl has at least two sweets?

how can i consider here that remaining 30 sweets can be repeated. i was considering here that there is 30 sweets remaining and for that we can select students in simply c(50,3) ways. but in answer it is c(79, 30). please explain.

user292174
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    Possible duplicate of Counting non-negative integral solutions – JMoravitz Nov 21 '16 at 05:10
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    After distributing the initial 70 sweets, we still have 30 more to distribute to the 50 people. The people as the bins and the sweets as the balls, we have $\binom{balls+bins-1}{bins-1}=\binom{30+50-1}{50-1}=\binom{79}{49}=\binom{79}{30}$ ways to do this. An answer of $\binom{50}{30}$ would have only counted those situations where people get at most one sweet extra. – JMoravitz Nov 21 '16 at 05:11
  • Presumably the hundred sweets are indistinguishable, but the boys and girls are considered distinct individuals. – hardmath Nov 21 '16 at 14:55

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First, give $1$ sweet to each boy and $2$ sweets to each girl. This uses up $70$ sweets, with $30$ left.

Then apply Stars-and-Bars to distribute the remaining $30$ sweets to the $50$ children. Number of ways is $$\binom {50+30-1}{30}=\binom{79}{30}$$

Hypergeometricx
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This is a classic "stars and bars" problem, allocating the remaining 30 (identical) sweets among the 50 children (after the pre-determined 70 sweets have been handed out).

In the normal statement of this you allocate dividers to separate out the 30 sweets between the children, so you need one fewer divider than the number of kids.

However it might be easier in this case to imagine a box with 79 slips of paper, with 30 saying "take a sweet" and 49 saying "pass the box". Line up the children, hand the first kid the box and then the order that those slips come out of the box determines who gets the sweets, with all possibilities possible. So to determine that order, choose the position of the 30 "take a sweet" slips in the 79 total slips of paper: ${79 \choose 30}$.

Joffan
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