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

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Question

(b) $P(n,r) = 1680$$\frac{{n!}}{{(n - r)!}} = 1680$&hellip;..$(i)$

$C(n,r) = 70$ ==>$\frac{{n!}}{{r!\,(\,n - r)!}} = 70$&hellip;..$(ii)$  $

Vedclass · Accepted Answer

$576$

Vedclass · Answer

(b) $P(n,r) = 1680$$\frac{{n!}}{{(n - r)!}} = 1680$&hellip;..$(i)$

$C(n,r) = 70$ ==>$\frac{{n!}}{{r!\,(\,n - r)!}} = 70$&hellip;..$(ii)$  $\frac{{1680}}{{r!}} = 70$, [From $(i)$ and $(ii)$]

$r! = \frac{{1680}}{{70}} = 24$==> $r = 4$

$\because $ $p(n,4) = \,1680$

$	herefore$ $n(n - 1)(n - 2)(n - 3) = 1680$ ==> $n = 8$

$8 	imes 7 	imes 6 	imes 5 = 1680$

Now $69n + r\,! = 69 	imes 8 + 4!$

$ = 552 + 24$ $= 576$.

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

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