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Question

Find the number of terms in the expansion of (2x+3y+z)7.

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Solution

Fact: Number of dissimilar terms in the expansion of (x1+x2+x3+......+xr)n is n+r1Cn

So in (2x+3y+z)7, we have r=3 and n=7
Thus number of terms in the expansion of (2x+3y+z)7 is 7+31C7=9C7=36

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