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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $2$).Since the tickets must be chosen from $15$ consecutively numbered tickets,we can treat the $10$ selected tickets as $10$ items and the $5$ unselected tickets as $5$ items.Total items = $15$. We arrange $10$ selected and $5$ unselected items such that the selected ones form blocks of $5, 3, 2$.This is equivalent to arranging $8$ items (where $5$ are the blocks and $3$ are the gaps/unselected items).The number of ways to arrange these $5$ blocks and $3$ gaps is $\frac{8!}{5!3!} = ^8C_5$.Since the $3$ children are distinct,the $3$ blocks can be distributed among them in $3!$ ways.Additionally,the internal arrangement of the blocks themselves relative to the children is $3!$.Thus,the total number of ways is $^8C_5 	imes 3! 	imes 3! = ^8C_5 	imes (3!)^2$.

The number of ways in which $3$ children can distribute $10$ tickets out of $15$ consecutively numbered tickets such that they get consecutive blocks of $5$,$3$,and $2$ tickets is:

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