There are $3$ black, $2$ green, and $1$ red ball in the basket. How many ways there is to pick up $4$ balls from that basket?

I know the answer is $5$. I solved this problem using generating function but I want to have a closed-form solution. Primarily I thought the answer is $$\frac{3+2+1\choose4}{3!\cdot2!\cdot1!}$$ but after writing down all possible sets of four balls, these two results were different. In fact, I don't understand why this way of thinking gives wrong result.

  • 2
    $\begingroup$ Not every selection is overcounted by $\binom{3+2+1}{4}$ in $3!2!1!$ ways. $\endgroup$ – Mauve Feb 15 '18 at 15:11
  • $\begingroup$ You can yourself check your mistake. Consider a similar question. How many 4 letter sets can be formed using 3 A's, 2 B's, and 1 C. $\endgroup$ – Rohan Shinde Feb 15 '18 at 15:17

Case 1) If 3 balls are alike:

Then number of ways = $\binom {2}{1}$

Case 2) If 2 balls are alike, and other two balls are also alike:

then number of ways =$\binom {2}{2}$

Case 3) If 2 balls are alike and other two balls are different :

Then number of ways =$\binom {2}{1}$

Hence total number of ways = $\binom {2}{1}+\binom {2}{2}+\binom {2}{1}=5$

  • $\begingroup$ The sum of binomials is 4 not 5. $\endgroup$ – mrJoe Feb 15 '18 at 15:24
  • $\begingroup$ @mrJoe Sorry of the typo. Answer edited. Thanks $\endgroup$ – Rohan Shinde Feb 15 '18 at 15:34

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.