We need to find the coefficient of x7 in (1−x+2x3)10.
We know that for (x+y+z)n,
Tn=a!b!c!n!(x)a(y)b(z)c such that a+b+c=n
Now for (1−x+2x3)10
Tn=a!b!c!10!(1)a(−x)b(2x3)c
=a!b!c!10!(−1)b2cxb+3c. with a+b+c=10
Here we need coefficient of x7.
Hence the combinations would be
a357b741c012
which satisfies both a+b+c=10 and b+3c=7.
Hence, coefficient of x7 will be,
=2!1!7!10!(2)2(−1)1
+1!4!5!10!(2)1(−1)4+0!7!3!10!(2)0(−1)7
=960