The number of ways in which 3 children can di | vedclass

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Question

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 distri

Vedclass · Accepted Answer

$^8C_5.(3!)^2$

Vedclass · Answer

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 $\frac{{8!}}{{5!}}3! = {\,^8}{{\rm{C}}_5} \cdot {(3!)^2}$

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

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