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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
$^8{C_5}$
B
$^8{C_5} 3!$
C
$^8{C_5} (3!)^2$
D
none of these
Solution
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 $3$ children i.e. $\frac{8 !}{5 !} 3 !$
Standard 11
Mathematics