# In how many ways can we pick three (3) balls so that we have a ball of each color?

1. We have a bin with 9 balls, 4 blue balls and 5 red balls. In how many ways can we pick three (3) balls so that we have a ball of each color?

2. We have a deck of 52 playing cards, 13 of each of the four suits. In how many ways can we pick five (5) cards so that there are at most two suits represented among the five cards? (As e.g. in three hearts and two clubs, but no spades and no diamonds.)

For the first question, I got 140 as my answer. I divided it into two cases: 1R 2B or 1B 2R, but i don't know if I did it correctly.

2((5C1)(4C2)) + 2((5C2)(4C1)) = 140

I am just getting started with this topic, can anyone help me with these questions, thanks.

• How did you calculate the numbers for 1R 2B and 1B 2R? – Théophile Mar 23 at 18:17
• It looks like you calculated the number of ways of selecting two blue balls and one red ball in that order or two red balls and one blue ball in that order. If all we care about is which balls are selected, the order of selection does not matter. Please edit your question to show your actual calculation so that we do not have to speculate about how you solved the problem. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Mar 23 at 18:18
• If you do care about order, then you should also have had the cases BRB and BBR as well as RBR and RRB in addition to the two cases you already looked at of RBB and BRR. – JMoravitz Mar 23 at 18:20
• I think we're not supposed to care about the order. So how would I solve it without caring about order? – rubik Mar 23 at 18:23
• Why did you include a factor of $2$ in each term? – N. F. Taussig Mar 23 at 18:27

For the first one, there are $$4$$ blue balls (B) and $$5$$ red balls (R). For the given condition, you choose either ($$1$$ B, $$2$$ R) or ($$2$$ B, $$1$$ R). The question does not specify that order matters so you should not multiply by $$2$$.

$$\displaystyle \small 4 \cdot {5 \choose 2} + 5 \cdot {4 \choose 2} = 70$$

For the second one, divide it into two cases, you either have all $$5$$ cards from one suit or from two suits. So first choose the suit(s) from where we pick $$5$$ cards and then choose $$5$$ cards either from $$13$$ (one suit) or $$26$$ (two suits)

So it should be $$\ \displaystyle \small 4 \cdot {13 \choose 5} + {4 \choose 2} \cdot \left[{26 \choose 5} - 2 \cdot {13 \choose 5}\right]$$.

$$\small \displaystyle {26 \choose 5} - 2 \cdot {13 \choose 5} \$$ is the number of ways in which we select $$5$$ cards such that it contains cards from both suits.

• I understand now, thank you! – rubik Mar 23 at 18:40
• you are welcome. – Math Lover Mar 23 at 18:42
• @rubik If Math Lover's answer resolves your questions, please accept the answer by clicking the check mark on the left, that is just below his score for this answer. – user2661923 Mar 23 at 20:07
• @user2661923 thank you! – Math Lover Mar 23 at 21:17