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

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(C) We have $15$ consecutively numbered tickets. We need to select $10$ tickets such that they form $3$ consecutive blocks of sizes $5, 3,$ and $2$.Le

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

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(C) We have $15$ consecutively numbered tickets. We need to select $10$ tickets such that they form $3$ consecutive blocks of sizes $5, 3,$ and $2$.Let the blocks be $B_1$ (size $5$),$B_2$ (size $3$),and $B_3$ (size $2$).To select these blocks from $15$ tickets,we can think of this as arranging $3$ blocks and $5$ empty spaces (since $15 - 10 = 5$ tickets are left out).Let the $5$ remaining tickets be represented by $X$. We need to arrange $B_1, B_2, B_3, X, X, X, X, X$.There are $8$ items in total to arrange,where $5$ items $(X)$ are identical.The number of ways to arrange these is $\frac{8!}{5!} = 8 	imes 7 	imes 6 = 336$.Since the $3$ children can be assigned these $3$ distinct blocks in $3!$ ways,the total number of ways is $\frac{8!}{5!} 	imes 3! = 336 	imes 6 = 2016$.Note that $\frac{8!}{5!} = ^8P_3 = 8 	imes 7 	imes 6 = 336$,and $^8C_5 = \frac{8!}{5!3!} = 56$.Thus,the answer is $336 	imes 6 = 2016$,which is $^8C_5 	imes 3! 	imes 3! = 56 	imes 6 	imes 6 = 2016$.

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

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