Q) A box contains $5$ red balls, $3$ blue balls, and $2$ yellow balls. Assume that all balls are different. In how many ways can you select $4$ balls so that each selection contains:

  1. Exactly two red balls$\quad\dots\space {^5C_2} \times {^5C_2}\quad?$

  2. At least two red balls?

  3. No yellow balls

  4. At most two red balls

  • $\begingroup$ Given the balls are different, you can think of this as the ten balls being numbered (red 1, red 2, ..., yellow 9, yellow 10). For example, question (1) you have correct: A selection of four balls which has exactly 2 red is $^5C_2 \cdot ^5C_2$. As there are $^5C_2$ choices for the two red balls, leaving 8 balls of which you can not choose red and therefore $^5C_2$ choices for your remaining two balls. Question (2) gives $^8P_2$ remaining choices as you can choose red. $\endgroup$
    – Nic
    Commented Sep 30, 2014 at 5:51

1 Answer 1


You are right with the first one

  1. C(5, 2) * C(5, 2)+C(5, 3)*C(5, 1)+C(5, 4)

You can select two red balls form the four red balls and two from the other remaining five balls. Similarly you can select three red balls and one front the remaining. Touch can also select four red balls.

  1. C(5, 4)

You can select any balls other than red.

4.C(5, 4)+ C(5, 1)*C(5, 3)+C(5, 2)*C(5, 2)

The selection can have no red balls , only one red ball or only two red balls

  • $\begingroup$ You misread the third question. It asks for he probability that no yellow balls are selected. $\endgroup$ Commented Mar 17, 2022 at 19:41

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