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I have 2 different combinations problems but I don't really understand how I would solve this. I know how to find probability but that is not what is asked for.

In a box, there are $8$ red and $6$ blue markers. How many ways can you select $3$ markers if:

(a) exactly $1$ red is selected?

(b) no more than $2$ blue are selected?

(c) there are more blue than red selected?

- A factory has $7$ female employees and $28$ male employees. The boss wants to create a party committee consisting of $5$ employees. In how many different ways can the boss select $5$ employees for the committee?

On this one I tried $\binom{28}{5}\binom{7}{5}$ and simply $\binom{35}{5}$, but I know neither answer is right.

N. F. Taussig
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Mark Gregg
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1 Answers1

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For part a, observe that there are ${8 \choose 1}=8$ ways to select the red marker. The remaining two markers have to be blue, so there is $6 \choose 2$ ways to select them. By the basic principle of counting, you can multiply these to obtain the total combinations: $$8{6 \choose 2}.$$

For part b, we need to add up the ways that there can be zero blue, one blue, and two blue. Additionally, we can more easily approach this problem by determining the ways to have all three blue, and subtracting that from the total possible combinations. So we have: $${14 \choose 3} - {6 \choose 3}.$$

For part c, there are only two possible configurations with more blue than red: three blues, or two blues and one red. So, we add the possible combinations corresponding to these: $${6 \choose 3} + 8{6 \choose 2}.$$

Let me know if any additional clarification is needed.

As Mark said, ${35 \choose 5}$ should be correct for the last question.

bwllmsn
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