A) 128
B) 576
C) 256
D) 625
E) 1152
Correct Answer: B
Solution :
\[\because \] \[^{n}{{p}_{r}}=1680\] \[\Rightarrow \] \[\frac{n!}{(n-r)!}=1680\] ...(i) and \[^{n}{{C}_{r}}=70\] \[\Rightarrow \] \[\frac{n!}{(n-r)!r!}=70\] ?.(ii) From Eqs. (i) and (ii), we get \[r!=\frac{1680}{70}=24=4!\Rightarrow r=4\] On putting value of r in Eq. (i), we get \[\frac{n!}{(n-4)!}=1680\] \[\Rightarrow \] \[n(n-1)(n-2)(n-3)=1680\] \[\Rightarrow \] \[n(n-1)(n-2)(n-3)=8(8-1)\] \[(8-2)(8-3)\] \[\Rightarrow \] \[n=8\] \[\therefore \] \[69n+r!=69\times 8+24\] \[=552+24+576\]You need to login to perform this action.
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