Question

If $15$ boys are randomly divided into $3$ groups of $4,5$ and $6$ boys, the probability that the $3$ youngest boys are in different groups is

• A. $12/91$
• B. $18/91$
• C. $24/91$
• D. $4/91$

Answer 1

Answer: (C)

$24/91$

we divide 3 youngest boys in 3 groups and remaining divided as follows $3!{\times }_3^{12}\text{C}{\times }_4^9\text{C}{\times }_5^5\text{C}$

division of 15 boys into 3 groups of 4,5,6 boys as follows ${}_4^{15}\text{C}{\times }_5^{11}\text{C}{\times }_6^6\text{C}$

Therefore probability of 3 youngest boys are in different group is=$\frac{3!{\times }_3^{12}\text{C}{\times }_4^9\text{C}{\times }_5^5\text{C}}{{}_4^{15}\text{C}{\times }_5^{11}\text{C}{\times }_6^6\text{C}}=\frac{24}{91}$