a, b, c are three particular speakers among t | vedclass

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(D) Total number of ways to arrange $10$ speakers in a row is $10!$. To find the number of ways where $a, b, c$ do not sit together,we subtract the nu

Vedclass · Accepted Answer

$84(8!)$

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(D) Total number of ways to arrange $10$ speakers in a row is $10!$. To find the number of ways where $a, b, c$ do not sit together,we subtract the number of ways where they do sit together from the total. Treating $a, b, c$ as a single unit,we have $8$ units to arrange (the group of $3$ and the remaining $7$ speakers),which can be done in $8!$ ways. Within the group,$a, b, c$ can be arranged in $3! = 6$ ways. So,the number of ways they sit together is $6 	imes 8!$. The number of ways they do not sit together is $10! - 6 	imes 8!$. $= (10 	imes 9 	imes 8!) - (6 	imes 8!) = (90 - 6) 	imes 8! = 84 	imes 8!$.

$a, b, c$ are three particular speakers among the $10$ speakers of a meeting. The number of ways of arranging all the $10$ speakers on the dais in a row so that all the three speakers $a, b, c$ do not sit together is

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