Question

If \[P(n,r) = 1680\] and, then find the value of \[69n + r!\] .
A.128
B.576
C.256
D.625
E.1125

Answer 1

Hint: First write the formula of \[P(n,r)\] and \[C(n,r)\]. Then divide \[P(n,r)\] by \[C(n,r)\] and obtain the value of r as 4. Then substitute the value of r in one of the given equations to obtain the value of n as 8. Substitute the values of n and r in the expression \[69n + r!\] and obtain the required value.

Formula used:
Formula of Permutation: \[P(n,r) = \dfrac{{n!}}{{\left( {n - r} \right)!}}\]
Formula of Combination: \[C(n,r) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]

Complete step by step solution:
It is given that,
\[P(n,r) = \dfrac{{n!}}{{\left( {n - r} \right)!}} = 1680 - - - - (1)\]
\[C(n,r) = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} = 70 - - - - (2)\]
Divide (1) by (2),
 \[\dfrac{{\dfrac{{n!}}{{\left( {n - r} \right)!}}}}{{\dfrac{{n!}}{{r!\left( {n - r} \right)!}}}} = \dfrac{{1680}}{{70}}\]
\[r! = 24\]
    \[ = 4!\]
\[\therefore r = 4\]
Substitute 4 for r in the equation (1) to obtain the value of n.
\[\dfrac{{n!}}{{\left( {n - 4} \right)!}} = 1680\]
\[n(n - 1)(n - 2)(n - 3) = 1680\]
\[n(n - 1)(n - 2)(n - 3) = 8 \times 7 \times 6 \times 5\]
Therefore, n=8
Now, put 8 for n and 4 for r in the expression \[69n + r!\] to obtain the value.
\[69 \times 8 + 4!\]
\[=552+24\]
\[=576\]

The correct option is B.

Note Sometime students stuck in the step \[n(n - 1)(n - 2)(n - 3) = 1680\] and unable to find the value of n, for that just factorise 1680 and find out the consecutive 4 numbers for multiplication.