There are five boxes consisting of several identical coins.

Question

There are five boxes consisting of several identical coins. The first box contains 4 coins, the second box contains 9 coins, the third box contains 7 coins, the fourth box contains 10 coins, and the fifth box contains 3 coins. Some coins are taken from the boxes such that at least a coin taken from each box. In how many ways in taking the coins, such that the numbers of the coins taken are 12?

My attempt was solving it using repetition combination of which, there are only seven coins remaining considering that at least a coin taken from each box. By using the repetition combination, i.e. taking 7 from 9 (five boxes and four borders), resulting in 36 ways. However, there is no 36 ways in the choices.

Answer 1

You will need to find the coefficient of $x^{12}$ in the series $(x+x^2+x^3+x^4)(x+x^2+x^3+...+x^9)(x+x^2+x^3+...+x^7)(x+x^2+x^3+...+x^{10})(x+x^2+x^3)$

i.e. the coefficient of $x^{7}$ in the series $(1+x+x^2+x^3)(1+x+x^2+x^3+...+x^8)(1+x+x^2+x^3+...+x^6)(1+x+x^2+x^3+...+x^8)(1+x+x^2)$

which can be further simplified to the coefficient of $x^{7}$ in the series $(1-x^4)(1-x^9)(1-x^7)(1-x^{10})(1-x^3)(1-x)^{-5}$

Well, it can be furthur simplified leaving all the power greater than 7 and power of sevel itself will cancel. $(1-x^3-x^4...)(1-x)^{-5}$

=$C(5+7-1,5-1)-C(5+4-1,5-1)-C(5+3-1,5-1)$

=$C(11,4)-C(8,4)-C(7,4)$

Answer 2

Num available: 4,9,7,10,3 - Num needed to take: 12

Take one from each box ahead of time to handle the condition that we need to take at least one from every box

Num available: 3,8,6,9,2 - Num needed to take: 7

Now... if we were to ignore the upper limits on each box, the number of ways to take 7 objects from 5 boxes would be:

$$\binom{7+5-1}{5-1}=330$$

You said something about $36$, which implies to me that you have made a mistake or learned stars and bars incorrectly. With $k$ distinct boxes and $n$ identical balls there are $\binom{n+k-1}{k-1}$ ways to place the balls in the boxes (or as in our case, remove balls from the boxes/coins who have an unlimited supply)

Of these outcomes however, you counted some impossibilities, such as where you took too many coins from the first box, so let us remove those impossibilities. If we took too many from the first box, that means we took at least four more from the first box, putting us at the situation of:

Num available: -1, 8,6,9,2 - Num needed to take: 3

Still five boxes, and needing to take only three more, there are $\binom{3+5-1}{5-1}=\binom{7}{4}=35$ bad outcomes where we took too many coins from the first box. We similarly count how many bad outcomes there were for having taken too many from the third or fifth box.

Note however, that in counting the number of bad outcomes as a result of taking too many from the first box and counting the number of bad outcomes as a result of taking too many from the fifth box, we accidentally counted one of these outcomes twice... the one where we took too many from both the first and fifth box simultaneously. Correctly applying inclusion-exclusion then we arrive at a final answer of:

$$\binom{7+5-1}{5-1}-\binom{3+5-1}{5-1}-\binom{0+5-1}{5-1}-\binom{4+5-1}{5-1}+\binom{0+5-1}{5-1}$$

$$ = \binom{11}{4}-\binom{7}{4}-\binom{4}{4}-\binom{8}{4}+\binom{4}{4} = 225$$

Answer 3

After taking one coin out of each box the boxes have contents $2,3,6,8,9$, and we have to pick $7$ more coins. From the first box we can take $0,1,2$ coins, and from the second box $0,1,2,3$ coins. This gives $1,2,3,3,2,1$ ways to take $0,1,2,3,4,5$ coins from these two boxes. The remaining $7,6,5,4,3,2$ coins to pick can be taken arbitrarily from the three other boxes, with one exception: you cannot take all $7$ coins from the $6$-coins box. Using stars and bars for these pickings we therefore obtain the following total number of admissible cases: $$N=1\cdot\left({9\choose2}-1\right)+2\cdot{8\choose2}+3\cdot{7\choose2}+3\cdot{6\choose2}+2\cdot{5\choose2}+1\cdot{4\choose2}=225\ .$$