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

combinationscombinatorics

  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.

Best Answer

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.

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