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Question

If 19Cr and 19Cr−1 are in the ratio 2:3, then find 14Cr

A
91
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B
81
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C
71
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D
61
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Solution

The correct option is A 91
We know that, nCr=n!r!(nr)!
Given, 19Cr19Cr1=23
=>19!r!(19r)!19!(r1)!(19r+1)!=23
=>(r1)!(20r)!r!(19r)!=23
=(r1)!×(20r)×(19r)!r×(r1)!×(19r)!=23
=>20rr=23
=>r=12
So,
14C12=14!12!(1412)!=14!12!(2)!=14×13×12!12!×2=7×13=91

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