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 from the lot.

Home » 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 from the lot.

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 from the lot.

Solution:

It is known that,

$\begin{array}{l} { }^{n} C_{r} \\ =\frac{n !}{r !(n-r) !} \end{array}$

As per the question,

5 = No. of black balls

6 = No. of red balls

${ }^{5} \mathrm{C}_{2}$ = No. of ways in which 2 black balls can be selected

$=\frac{5 !}{2 ! 3 !}=10$

${ }^{5} \mathrm{C}_{3}$ = No. of ways in which 3 red balls can be selected

$=\frac{6 !}{3 ! 3 !}=20$

Total no. of ways in which 2 black and 3 red ball can be selected

$\begin{array}{l} ={ }^{5} \mathrm{C}_{2} \times{ }^{5} \mathrm{C}_{3} \\ =10 \times 20 \\ =200 \end{array}$

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