Calculating combinations where all the elements chosen are the same.

Question

If I have 7 red balls and 8 blue balls and have to choose 4 balls from this 15, how many possible combinations can I get where all 4 balls are red balls?

I've tried using the Combinations formulae where I'm basically selecting 4 elements from the 7 [of the red balls] and coming away with: $$\displaystyle\frac{7!}{(7-4)! 4!}$$ where there are 35 possible combinations, although this doesn't seem correct to me. Am I on the right track with this line of thinking?

I think I was second guessing myself to be honest. Thanks for the clarification! — Broadsword93, Aug 20 '17 at 14:50

score 1 · Accepted Answer · answered Aug 21 '17 at 08:16

You are correct.

The number of ways of selecting exactly $k$ of the $7$ red balls and $4 - k$ of the $8$ blue balls when $4$ balls are chosen is $$\binom{7}{k}\binom{8}{4 - k}$$ so the number of ways of selecting $4$ red balls is $$\binom{7}{4}\binom{8}{0} = \binom{7}{4}$$

Calculating combinations where all the elements chosen are the same.

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