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Bunuel
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Answer should be B substitute 3 5 and 7 to find out.
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nC1=n!/(n-1)!=n
nC2=n!/(2*(n-2)!)=n*(n-1)/2

(A): n/nC1= n/n=1 OK
n/nC2=2(n-1), and since n is odd integer greater than 2, n>3. 2/(3-1)=1, 2/(5-1)=1/2 NOT OK.

(B) n(n-1)/2/nC1=(n-1)/2 OK
n(n-1)/2/nC2=1 OK

(C) doesn't work

(D) doesn't work

(E) doesn't work
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nC1= n and nC2=n(n - 1)/2
We see that nC2 is a multiple of nC1.
The LCM of two numbers where one is the multiple of the other, is the larger number.
For eg: The LCM of 12 and 48 is 48.
We know that (n - 1)/2 is greater than 1 as n is greater than 2. Therefore, n(n - 1)/2 is greater than n and hence the answer.
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Bunuel
What is the LCM of nC1 and nC2, where n is an odd integer greater than 2?

(A) n

(B) n(n - 1)/2

(C) (n - 1)/2

(D) n/2

(E) n − 1

nC1 = n! /(n-1)!*1! = n
nC2 = n! /(n-2)!*2! = n (n-1)/2

We know that n is odd and greater than 2, thus n-1 is even (cannot be 0). Thus n-1 is divisible by 2 and will lead to an integer.
LCM = n (n-1)/2 (B)
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