Question

The number of ways in which 3 children can distribute 10 tickets out of 15 consecutively numbered tickets among themselves such that they get consecutive blocks of 5, 3 and 2 tickets is

• A. ${}^8{C}_3$
• B. ${}^8{C}_3\times 3!$
• C. ${}^8{C}_3\times (3!{)}^2$
• D. $Noneofthese$

Answer 1

The correct option is C

${}^8{C}_3\times (3!{)}^2$

Problem is same as arranging 8 things out of which 5 identical i.e. $\frac{8!}{5!}$ which gives total number of ways of selecting block and distributing them away to 3 children $=\frac{8!}{5!}\times 3!$