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

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6.Permutation and Combination

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

A

$128$

B

$576$

C

$256$

D

$625$

Solution

(b) $P(n,r) = 1680$$\frac{{n!}}{{(n – r)!}} = 1680$…..$(i)$

$C(n,r) = 70$ ==>$\frac{{n!}}{{r!\,(\,n – r)!}} = 70$…..$(ii)$ $\frac{{1680}}{{r!}} = 70$, [From $(i)$ and $(ii)$]

$r! = \frac{{1680}}{{70}} = 24$==> $r = 4$

$\because $ $p(n,4) = \,1680$

$\therefore$ $n(n – 1)(n – 2)(n – 3) = 1680$ ==> $n = 8$

$8 \times 7 \times 6 \times 5 = 1680$

Now $69n + r\,! = 69 \times 8 + 4!$

$ = 552 + 24$ $= 576$.

Standard 11

Mathematics

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