First slide

Combinations

Question

In how many ways can a team of 6 horses be selected out of a stud of 16, so that there shall always be three out of A B C A' B' C', but never A A' , B B' or C C' together

Moderate

A

840
B

1260
C

960
D

720

Solution

The number of ways is $^{10} C_{3} \times$ number of ways of choosing out of ABC A'B'C', so that AA', BB' or CC' are not together
$\begin{array}{l} =^{10} C_{3} (one from each of pairs {AA}^{'}, {BB}^{'}, {CC}^{'}) \\ =^{10} C_{3} \times 8 \\ = \frac{10 \times 9 \times 8}{1 \times 2 \times 3} \times 8 = 960 \end{array}$

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