2
$\begingroup$

Given a multiset that contains 5 numbers where the numbers are from 0 to 5 inclusive, and the numbers can be repeated:

a) In how many ways can you have a multiset with exactly four 4s?

b) In how many ways can you have a multiset with exactly three 3s?

c) In how many ways can you have a multiset with exactly two 2s?

d) In how many ways can you have a multiset with exactly one 1s?

e) In how many ways can you have a multiset with exactly zero 0s?

I know that the number of repeated combinations of $k$ elements from $n$ available elements is given by the expression ${n+k-1 \choose k}$, but how would you apply it to this problem.

For example, for a) would it be ${5+3-1 \choose 3} = 35$

I am very confused on this subject, and require some help.

$\endgroup$
1
$\begingroup$

We do exactly two $2$'s, you can do the rest.

We need to count the number of ways to produce "the rest" of the multiset. So we need to count the number of $3$-element multisets, where the elements are chosen from the collection $\{0,1,3,4,5\}$.

You quoted a formula for counting the number of $3$-element multisets taken from a collection of $5$ numbers. It yields the answer $\binom{5+3-1}{3}$.

Remark: We certainly do not need the formula to deal with Question (a). For we need to choose one non-$4$ to go with the four $4$'s. There are $5$ numbers ot choose from, so we have $5$ choices.

The formula you mentioned does work, however, with $n=5$ and $k=1$.

$\endgroup$
  • $\begingroup$ That makes much more sense now, thanks. $\endgroup$ – Vishwa Iyer Jul 14 '13 at 2:28
  • $\begingroup$ You are welcome. I have added a remark about your answer for (a). $\endgroup$ – André Nicolas Jul 14 '13 at 2:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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