Find the value of r: 11p3r=110
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Class 11
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>>Applications of Permutations and Combinations
>>If P(n,r) = 2520 and C(n,r) = 21 , then
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If P(n,r)=2520 and C(n,r)=21, then what is the value of C(n+1,r+1)?
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Correct option is C)
Given,
P(n,r)=2520
(n−r)!
n!
=2520...(1)
C(n,r)=21
r!(n−r)!
n!
=21....(2)
comparing (1) and (2), we get,
r!
2520
=21
∴r!=120
r=5
Now,
P(n,5)=2520
(n−5)!
n!
=2520
n(n−1)(n−2)(n−3)(n−4)=2520
n=7
C(n+1,r+1)
=C(7+1,5+1)
=C(8,6)
=
2
8×7
=28
Step-by-step explanation:
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