1+C1C01+C2C11+C3C2……1+CnCn-1=
[n+1]n!
[n+1]n(n–1)!
[n+1]nn!
[n–1]nn!
Explanation for the correct option:
Step 1. Expand C0and C1:
Given, 1+C1C01+C2C11+C3C2……1+CnCn-1
As we know,
C1n=n!(n−1)!1!=n ∵Crn=n!(n−r)!r!
C0n=n!(n−0)!0!=1
Step 2. Divide C1n by C0n, we get
C1C0=n
Similarly,
C2C1=n!n-2!2!n!n-1!1!=n-1n-2!n-2!2=n-12
C3C2=n!n-3!3!n!n-2!2!=n-2n-3!2n-3!3×2=n-23
CnCn-1=n!n-n!n!n!n-n+1!n-1!=n-n+1n-1!n-n!n!=n-1!n×n-1!=1n
Step 3. Put all these values in given expression:
1+C1C01+C2C11+C3C2……1+CnCn-1=1+n1+n-121+n-23...1+1n
=1+n11+n21+n3...1+nn
=(1+n)nn! ∵1.2.3.....n=n!
Hence, Option ‘C’ is Correct.