Example 1: If 15 boys of different ages are distributed into 3 groups of 4, 5 and 6 boys randomly - Maths - Probability

Question

Example 1: If 15 boys of different ages are distributed into 3
groups of 4, 5 and 6 boys randomly then the probability that three
youngest boys are in different groups is
(a) 24 91
(b) 71 91
(c) 67 91
(d) 20 91

Shruti Tyagi · Accepted Answer

Dear Student,

Total Number of ways of distributing 15 boys in 3 groupsof 4,5 and 6 
first select 4 boys out of 15, now from remaining 11 boys select any 5 and than remaining 6 in third group
=C415×C511×C66=15!11!×4!×11!6!×5!×6!6!×0!=15!4!5!6!The number of ways of distributing 12 remaining boys excluding the 3 youngest=C312×C49×C55=12!9!×3!×9!4!×5!×5!5!×0!=12!3!×4!×5!and number of ways of distributing 3 youngest boys is P33=3!The required Probabilility=3!×12!3!×4!×5!15!4!5!6!=12!×4!×5!×6!4!×5!×15!=12!×6!15!=6!15×14×13=6×5×4×3×215×14×13=2491

Regards,

Example.1: If 15 boys of different ages are distributed into 3 groups of 4, 5 and 6 boys randomly then the probability that three youngest boys are in different groups is (a) 24 91 (b) 71 91 (c) 67 91 (d) 20 91

Example.1: If 15 boys of different ages are distributed into 3 groups of 4, 5 and 6 boys randomly then the probability that three youngest boys are in different groups is
(a) $\frac{24}{91}$
(b) $\frac{71}{91}$
(c) $\frac{67}{91}$
(d) $\frac{20}{91}$