A bag contains 5 blue balls and 5 black balls. The number of ways in which a blue ball and a black ball can be taken out from the bag at once is

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Solution

A bag contains 5 blue balls and 5 black balls.

So, the number of ways in which a blue ball can be taken out
$=^{5} C_{1} = \frac{5!}{(5 - 1)! \times 1!}$
$= \frac{5!}{4! \times 1} = \frac{5 \times 4!}{4!} = 5$

And the number of ways in which a black ball can be taken out
$=^{5} C_{1} = \frac{5!}{(5 - 1)! \times 1!}$
$= \frac{5!}{4! \times 1} = \frac{5 \times 4!}{4!} = 5$

∴ Number of ways of taking out a blue ball and black ball = 5 × 5 = 25

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