Question

In the expansion of (1+x)n(1+y)n(1+z)n, the sum of coefficients of the terms of degree ‘r’ is

3nCr

3.nCr

3nC3r

3nC3n−r

Solution

The correct options are

**A**

**D**

3nCr

3nC3n−r

(1+x)n(1+y)n(1+z)n(∑nm=0 nCmxm)(∑ns=0 nCsys)(∑nt=0 nCtzt)=∑0≤r, s, t≤n nCm nCs nCt XmYsZt

For sum of the coefficients of degree ‘r’, we have m + s + t = r; where, m, s, t are integers greater than or equal to zero. Sum of such coefficients

=∑m,s,t≥0m+s+t=rnCm n Cs nCt (this is similar to the number of ways of choosing a total of r balls out of n black, n white, n green balls.) =3nCr

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