Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

Please Correct Me

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

share|cite|improve this question
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
up vote 1 down vote accepted

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.

share|cite|improve this answer
Thanks Man , Great Explanation, Thank you. – Node.JS Nov 21 '12 at 21:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.