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

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$^8C_5 \cdot (3!)^2$

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(C) We have $15$ consecutively numbered tickets. We need to select $10$ tickets such that they form blocks of sizes $5, 3,$ and $2$.Let the blocks be $B_1$ (size $5$),$B_2$ (size $3$),and $B_3$ (size $2$).Since the tickets must be chosen from $15$ consecutively numbered tickets,we are essentially arranging $3$ blocks and $5$ empty spaces (since $15 - 10 = 5$ tickets are left out).This is equivalent to arranging $8$ items (the $3$ blocks and $5$ gaps) in a row,which can be done in $\frac{8!}{5!}$ ways.After selecting the positions for the blocks,we assign the $3$ blocks to the $3$ children,which can be done in $3!$ ways.However,the problem implies the blocks are distinct by size. The number of ways to arrange the blocks is $\frac{8!}{5!} = ^8P_3 = ^8C_3 \cdot 3! = ^8C_5 \cdot 3!$.Since the children are distinct,we multiply by $3!$ for the distribution among them.Thus,the total number of ways is $\frac{8!}{5!} \cdot 3! = ^8C_5 \cdot 3! \cdot 3! = ^8C_5 \cdot (3!)^2$.

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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