If P(n,r) = 1680 and C(n,r) = 70,then 69n + r | vedclass

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(B) We know that $P(n,r) = \frac{n!}{(n-r)!} = 1680$ $(i)$ and $C(n,r) = \frac{n!}{r!(n-r)!} = 70$ $(ii)$.Dividing $(i)$ by $(ii)$,we get $\frac{P(n,r

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$576$

Vedclass · Answer

(B) We know that $P(n,r) = \frac{n!}{(n-r)!} = 1680$ $(i)$ and $C(n,r) = \frac{n!}{r!(n-r)!} = 70$ $(ii)$.Dividing $(i)$ by $(ii)$,we get $\frac{P(n,r)}{C(n,r)} = r! = \frac{1680}{70} = 24$.Since $r! = 24 = 4!$,we have $r = 4$.Substituting $r = 4$ into $P(n,4) = 1680$,we get $n(n-1)(n-2)(n-3) = 1680$.Testing values,$8 	imes 7 	imes 6 	imes 5 = 1680$,so $n = 8$.Finally,$69n + r! = 69(8) + 4! = 552 + 24 = 576$.

If $P(n,r) = 1680$ and $C(n,r) = 70$,then $69n + r! = $

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