# Probability, Finding number of cases

Suppose $x_1$ , $x_2$, $x_3$, $x_4$ >0 then $x_1$+$x_2$+$x_3$+$x_4$=18

a) $x_1$>5

b) $x_4$ $\neq$ 5

c) $x_2$=2 and $x_3$=5

d) $x_1$+$x_2$=5

a) $\binom{4+(18-6)-1}{18-6}$

b) $\binom{4+(18-6)-1}{18-6}$ - $\binom{4+(18-5-6)-1}{18-5-6}$

c) $\binom{2+11-1}{11}$

d) $\binom{2+5-1}{5}$*$\binom{2+13-1}{13}$

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I take it (c) asks for $x_2=2$ and $x_3=5$. That leaves $11$ between the other two guys. Since the $x_i$ have to be positive, there are $10$ ways to do it. If we allowed $0$, there would be $12$ ways, which is your answer. So if they really want positive, you are using wrong formula. Similar issue in (d) and probably elsewhere. –  André Nicolas Nov 21 '12 at 20:31
In my view such "questions" should not be accepted. In their refusal of producing even a measly representation of the actual problem they show a total disrespect for the community at work here. Giving an answer beginning with "I conjecture you are asking whether $\ldots$" is subsidizing such an attitude. –  Christian Blatter Nov 21 '12 at 20:57
@Hooman: Your question said explicitly $x_1,x_2,x_3,x_4 \gt 0$, and I analyzed and wrote up things, in fair detail, on that basis. It certainly should be not too difficult to proofread questions to see whether they accurately represent the problem. –  André Nicolas Nov 21 '12 at 21:15

The formulas being used are not quite correct. Note that the problem (at least as first and currently stated) says that the $x_i$ are bigger than $0$.

a) Give $5$ marbles to Kid $1$. We have $13$ marbles left, to be distributed between $4$ kids, at least one to each. By the usual Stars and Bars argument, we need to choose $3$ gaps from the $12$ available to put a separator into. There are $\binom{12}{3}$ ways to do this. Another kind of analysis will yield the equivalent answer $\binom{12}{9}$.

You have $\binom{15}{12}$, which is not equal to $\binom{12}{3}$.

(b) You used the right strategy, count all ways, subtract the ways in which $x_4=5$. Let's count these excluded cases. So we need to distribute $13$ marbles between $3$ kids, at least $1$ to each kid. There are $\binom{12}{2}=\binom{12}{10}$ ways to do this. You have something else. The first term is also not right, it should be $\binom{18}{3}$ or $\binom{18}{15}$. The issue here, as usual, is forgetting about the condition $x_i\gt 0$.

c) Now we have $11$ marbles, to be distributed between Kid $1$ and Kid $4$. Hardly need formulas for this. Since each kid has to get at least $1$ marble, there are $10$ ways to do this. You got $12$, which would be the right answer if we allow the possibility of a kid getting no marbles. However, that is ruled out by the beginning of the problem, which says $x_i\gt 0$.

d) There are $4$ ways to distribute $5$ marbles between Kids $1$ and $2$, at least $1$ to each kid. That leaves $13$ marbles to be distributed between Kids $3$ and $4$, at least one to each kid. There are $12$ ways to do this, for a total of $(4)(12)$. You got $(6)(14)$, which would be correct if we allow a kid to have no marbles. But that is excluded.

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Thanks Man , Great Explanation, Thank you. –  Hooman Nov 21 '12 at 21:40