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Question

If the number of distinct terms in expansion of (x+y+z+1xy+1yz+1zx)2 and (x+y+z+1x+1y+1z)2 is m and n respectively, then the value of m+n is

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Solution

We have,
(x+y+z+1xy+1yz+1zx)2
Total number of terms
= 2+61C61= 7C5=21
Similarly for
(x+y+z+1x+1y+1z)2
The total number of terms =21

Now, for first expansion, we get
1x=y×1xy or z×1zx1y=x×1xy or z×1zy1z=x×1xz or y×1zy
So, the number of distinct terms m=213=18

And, for the second expansion, we get
1=x×1x or y×1y or z×1z
So, the number of distinct terms n=212=19

Hence, m+n=18+19=37

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