Question

If 7 . 5pr = 7pr – 1, then find r.(question based on permutation and combination chapter)​

Answer 1

SOLUTION

TO DETERMINE

The value of r when

$\sf{ 7.{ \: }^{5}P_r = { \: }^{7}P_{r - 1} \: }$

EVALUATION

Here

$\sf{ 7.{ \: }^{5}P_r = { \: }^{7}P_{r - 1} \: }$

$\implies \displaystyle \sf{7 \times \frac{5! }{(5 - r)! } = \frac{7! }{(8 - r)! \: } }$

$\implies \displaystyle \sf{7 \times \frac{5! }{(5 - r)! } = \frac{7 \times 6 \times 5! }{(8 - r)(7- r)(6 -r )(5 -r )! } }$

$\implies \displaystyle \sf{7 = \frac{7 \times 6 }{(8 - r)(7- r)(6 -r ) } }$

$\implies \displaystyle \sf{(8 - r)(7- r)(6 -r ) = 6 }$

$\implies \displaystyle \sf{(8 - r)(7- r)(6 -r ) = 3 \times 2 \times 1 }$

Comparing both sides we get

$\sf{ 8 - r = 3\: }$

$\implies \sf{r = 5}$

FINAL ANSWER

Hence the required value of r is 5

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