- Home
- Standard 11
- Mathematics
6.Permutation and Combination
hard
જો $P(n, r) = 1680$ અને $C (n, r) = 70,$ હોય, તો $69 n + r! = ……$.
A
$128$
B
$576$
C
$256$
D
$625$
Solution
$\,P(n,r) = 1680$ $\frac{{n!}}{{(n – r)!}} = 1680\,\,\,\,………\,\,(i)$
$C(n,r) = 70\,\, \Rightarrow \,\,\frac{{n!}}{{r!\,(\,n – r)!}} = 70\,\,\,{\text{ }}..{\text{(ii) }}$
$\therefore \,\,\,\frac{{1680}}{{r!}} = 70\,{\text{, }}$
$r! = \frac{{1680}}{{70}} = 24\,\, \Rightarrow \,r = 4\,\,\,\,\,$
$P(n,\,4) = 1680\,\,\,\,\,$
$n(n – 1)(n – 2)(n – 3) = 1680$
$8 \times 7 \times 6 \times 5 = 1680$
હવે ,$69n + r\,! = 69 \times 8 + 4!\,\,\, = \,\,552\,\, + \,\,24\,\, = \,\,576.$
Standard 11
Mathematics
