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Question

If nPr=360 and nCr=15 then find the value of r

A
5
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B
4
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C
3
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D
2
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Solution

The correct option is D 4
nPr=n!(nr)!=360-- (1)
nCr=n!(r!)(nr)!=15 -- (2)
Dividing eqn 1 by 2, we get
=>r!=36015=24
=>r!=4×2×3×1=4!
So, r=4

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