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If \ ^(a^2-a)C2=\ \ ^(a^2-a)C4\ t h e n=...

If `\ ^(a^2-a)C_2=\ \ ^(a^2-a)C_4\ t h e n=` `2` b. `3` c. `4` d. none of these

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`\ ^n C_r=\ ^n C_y`
So,
`2+4=a^2-a`
`6=a^2 -a`
Now by solving with factorization we get `a=-3 or x=2`
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