- Home
- Standard 11
- Mathematics
6.Permutation and Combination
normal
The number of ways in which $3$ children can distribute $10$ tickets out of $15$ consecutively numbered tickets themselves such that they get consecutive blocks of $5, 3 $ and $2$ tickets is
A
$^8C_5$
B
$^8C_5.3!$
C
$^8C_5.(3!)^2$
D
$^{15}{C_{10}}.3!$
Solution
Problem is same as arranging $8$ things out of which $5$ identical ie $\frac{8 !}{5 !}$ which gives total number of ways of selecting block and distributing them away to $3$ children is $\frac{{8!}}{{5!}}3! = {\,^8}{{\rm{C}}_5} \cdot {(3!)^2}$
Standard 11
Mathematics
