MCQ - 149-8881

Correct Answer: C

Explanation:

8. Total number of persons = 6 + 3 = 9 Total number of equally likely cases = ⁹P₉ = 9 !

Let A denote the event that no two girls sit together.

First we place six boys at the six places B₁ , B₂ , …, B₆ . This can be done in 6 ! ways. X B₁ X B₂ X B₃ X B₄ X B₅ X B₆ X Since no two girls sit together,

we can seat them at X marked places.

No. of X marked places = 7

No. of ways for girls = ⁷P₃

total number of favorable cases = ⁷P₃ × 6 !

required probability = $\genfrac{}{}{1px}{}{\mathrm{7\; \times \; 6\; \times \; 5\; \times \; 6!}}{\mathrm{9\; \times \; 8\; \times \; 7\; \times \; 6!}}$

$\genfrac{}{}{1px}{}{\mathrm{6\; \times \; 5}}{\mathrm{9\; \times \; 8}}$ = $\genfrac{}{}{1px}{}{5}{\mathrm{3\; \times \; 4}}$ = $\genfrac{}{}{1px}{}{5}{\mathrm{12}}$