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

1) In how many ways can 10 pencils (identical) be distributed among 5 children if a) there are no restrictions? b) each child gets at least 1 pencil? c) the oldest child gets at least 2 pencils?

2) How many arrangements of the letters in MISSISSIPPI have no consecutive S' s?

3) In how many ways can a gambler draw 5 cards from a standard deck and get a) a flush (five cards of the same suit) ? b) four aces? c) four of a kind? d) three aces and two jacks e) three aces and a pair? f) a full house (three of a kind and a pair) ? g) three of a kind? h) two pairs?

Please help me to understand this topic because I am very new at this. If you solve them please explain them in some detail or steps. Thanks in advance.

share|cite|improve this question
@Kannappan: Your link has been Google-mangled. The full set of notes in updated form is at the link Course Notes here. – Brian M. Scott Feb 21 '12 at 23:56
@Brian Thanks for posting. I have deleted my previous comment. – user21436 Feb 22 '12 at 0:00

There are too many questions. I will deal with the card questions, since they form a connected collection.

(a) The suit (one of $\spadesuit$, $\heartsuit$, $\diamondsuit$, $\clubsuit$) can be chosen in $4$ ways. Instead, call this $\binom{4}{1}$. For every choice of suit, the actual $5$ cards can be chosen in $\binom{13}{5}$ ways, for a total of $\binom{4}{1}\binom{13}{5}$. I assume you know how to compute from this point on. Please note that our count included the straight flushes, and the royal flushes, which in poker are much better hands than a plain flush. If we want to just count the plain flushes, we need to subtract the number of straight flushes. But we were not asked to count the plain flushes.

(b) The aces can be chosen in only one way, though one could call the number $\binom{4}{4}$. The remaining card can be chosen in $48$ ways. So the total is $48$, but we could call that $\binom{4}{4}\binom{48}{1}$.

(c) We need to choose the kind. There are $\binom{13}{1}$ ways of doing this. For each such choice, there are $\binom{48}{1}$ ways to choose the remaining card, for a total of $\binom{13}{1}\binom{48}{1}$.

(d) There are $\binom{4}{3}$ ways to choose the three Aces. For each such choice, there are $\binom{4}{2}$ ways to choose the Jacks, for a total of $\binom{4}{3}\binom{4}{2}$.


(f) There are $\binom{13}{1}$ ways to choose the kind we will have three of. For each such choice, there are $\binom{4}{3}$ ways to choose the actual three cards. For every way of doing these two things, there are $\binom{12}{1}$ ways to choose the kind we will have two of, and $\binom{4}{2}$ ways to choose the actual cards, for a total of $\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}$.

(g) Like in (f), there are $\binom{13}{1}\binom{4}{3}$ ways to choose the kind we will have three of, and the actual cards. There remain two cards to choose. It is tempting to think that there are $\binom{48}{2}$ ways to choose these. (Of course we have to avoid the last card of the kind we have three of.) So it is tempting to think that the total is $\binom{13}{1}\binom{4}{3}\binom{48}{2}$. However, this would count the full house hands, which are a much better hand. So the actual total is $\binom{13}{1}\binom{4}{3}\binom{48}{2}-\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}$. We can also express this as $\binom{13}{1}\binom{4}{3}\left(\binom{48}{2}-\binom{12}{1}\binom{4}{2}\right)$.

(h) This one is quite tricky. It is all too easy to give a plausible argument that leads to an answer that is wrong by a factor of $2$. There are $\binom{13}{2}$ ways to choose the two kinds that we will have two cards from. For each such choice, there are $\binom{4}{2}$ ways to choose the actual two cards of the higher ranking kind that we chose, and then $\binom{4}{2}$ ways to choose the two cards of the lower ranking kind. Once this has been done, there are $\binom{44}{1}$ ways to choose the "odd" card, for a total of $\binom{13}{2}\binom{4}{2}\binom{4}{2}\binom{44}{1}$.

share|cite|improve this answer
I would just like to add that four of a kind means 2, 2, 2, 2 or 3, 3, 3, 3 or 4, 4, 4, 4 and so on. I thought it meant four of the same color so it had me confused for a while. – pug Dec 20 '12 at 10:10
g. Can also be solved as 13 * C(4,3) ways to choose three of a kind. Then we have two cards remaining. We can choose the first in 48 ways and then the second in 44 ways to avoid getting pairs. But now we have counted for example 2,3 as well as 3,2 however they are same for our purposes so we divide them by two resulting in 13 * C(4, 3) * 48 * 44 / 2. – pug Dec 20 '12 at 10:44
h. The answer should have C(44,1) instead of C(48,1) because we don't count the pairs we have already chosen. This makes eight cards in total. So, if we have 2, 2 and 3, 3 then we need to subtract 2,2,2,2 and 3,3,3,3 from the total cards to select the last one. Hence 52-8=44. – pug Dec 20 '12 at 11:37
@pug: Thanks, corrected. – André Nicolas Dec 20 '12 at 16:17

Here is a solution to your MISSISSIPPI problem.

Start with a string of seven stars for the letters not equal to "S". Now drop the four "S"s into the spaces between these stars, including one or both endpoints; for example: $S\,*\,*\,*\,S\,*\,S\,*\,*\,S\,*$. There are ${8\choose 4}=70$ ways to make this choice.

Now that we know where the "S"s go, let's fill in the seven stars with the remaining letters. There are ${7!\over 4! 2!}=105$ distinct ways to fill those places.

The total number of arrangements with non-consecutive "S"s is $70\cdot 105 =7350$.

share|cite|improve this answer

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.