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

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(D) Let the $15$ tickets be represented as $T_1, T_2, \dots, T_{15}$.We need to select $10$ tickets such that they form consecutive blocks of sizes $5

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none of these

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(D) Let the $15$ tickets be represented as $T_1, T_2, \dots, T_{15}$.We need to select $10$ tickets such that they form 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 treat the $5$ remaining tickets as separators.There are $15 - 10 = 5$ tickets that are not selected.Let these $5$ tickets be $X_1, X_2, X_3, X_4, X_5$.These $5$ tickets create $6$ possible gaps (including ends) where the blocks can be placed: $\_ X_1 \_ X_2 \_ X_3 \_ X_4 \_ X_5 \_$.We need to place $3$ distinct blocks $(B_1, B_2, B_3)$ into these $6$ gaps.The number of ways to choose $3$ gaps out of $6$ is $^6C_3$.Since the blocks are distinct $(5, 3, 2)$,they can be arranged in the $3$ chosen gaps in $3!$ ways.Total ways = $^6C_3 	imes 3! = 20 	imes 6 = 120$.Alternatively,if we consider the selection of the starting positions of the blocks such that they do not overlap,the calculation leads to $120$. None of the provided options match this result.

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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