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From three sets of twins and four unrelated girls, find how many selections of five people can be made if exactly:

a) two sets of twins must be included
b) one set of twins must be included.

What I did was

${}^3C_2 \times {}^4C_1 = 12$ and ${}^3C_2 \times {}^4C_3 = 12$

But the correct answer is 18 and 132.

Am I miss understanding the order of selection that it matters or something else?

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CasperYC
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1 Answers1

3

a)

You overlooked that it is also permitted to include a single member of one of the twins.

That means that your $4$ should be $4+2=6$:

$$\binom32\times\binom61=18$$

b)

After choosing one twin there are $3$ ways to arrive at $5$ persons:

  • $3$ of the $4$ girls are added.
  • $1$ member of a twin is added together with $2$ girls.
  • $2$ members of a twin are added that do not come from the same twin together with one girl.

$$\binom31\left[\binom43+\binom41\binom42+\binom21\binom21\binom41\right]=132$$

drhab
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