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

We have, ( x + y + z + 1xy + 1yz + 1zx )^2 Total number of terms = ^2+6 1C_6 1= ^7C_5=21 nbsp; Similarly for nbsp; nbsp; ( x + y + z + 1x + 1y + 1z )^2 T ...

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Byju's Answer

Standard XII

Mathematics

Multinomial Expansion

If the number...

Question

If the number of distinct terms in expansion of ${(x + y + z + \frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x})}^{2}$ and ${(x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z})}^{2}$ is $m$ and $n$ respectively, then the value of $m + n$ is

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Solution

We have,
${(x + y + z + \frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x})}^{2}$
Total number of terms
$=^{2 + 6 - 1} C_{6 - 1} =^{7} C_{5} = 21$
Similarly for
${(x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z})}^{2}$
The total number of terms $= 21$

Now, for first expansion, we get
$\frac{1}{x} = y \times \frac{1}{x y} or z \times \frac{1}{z x} \frac{1}{y} = x \times \frac{1}{x y} or z \times \frac{1}{z y} \frac{1}{z} = x \times \frac{1}{x z} or y \times \frac{1}{z y}$
So, the number of distinct terms $m = 21 - 3 = 18$

And, for the second expansion, we get
$1 = x \times \frac{1}{x} or y \times \frac{1}{y} or z \times \frac{1}{z}$
So, the number of distinct terms $n = 21 - 2 = 19$

Hence, $m + n = 18 + 19 = 37$

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Similar questions

Q. If the number of distinct terms in expansion of

{(x + y + z + \frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x})}^{2}

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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

If the number of distinct terms in expansion of ${(x + y + z + \frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x})}^{2}$ and ${(x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z})}^{2}$ is $m$ and $n$ respectively, then the value of $m + n$ is