A bag contains $5$ black and $6$ red balls. Determine the number of ways in which $2$ black and $3$ red balls can be selected.
A bag contains $5$ black and $6$ red balls. Determine the number of ways in which $2$ black and $3$ red balls can be selected.
There are $5$ black and $6$ red balls in the bag.
$2$ black balls can be selected out of $5$ black balls in $^{5} C_{2}$ ways and $3$ red balls can be selected out of $6$ red balls in $^{6} C_{3}$ ways.
Thus, by multiplication principle, required number of ways of selecting $2$ black and $3$ red balls
$=^{5} C_{2} \times^{6} C_{3}=\frac{5 !}{2 ! 3 !} \times \frac{6 !}{3 ! 3 !}=\frac{5 \times 4}{2} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1}=10 \times 20=200$
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- [KVPY 2019]