Question

$\left(^{8}C_{1} - ^{8}C_{2} + ^{8}C_{3} -^{8}C_{4} + ^{8}C_{5} - ^{8}C_{6} +^{8}C_{7} - ^{8}C_{8}\right) $ equals:

• A. 0
• B. 1
• C. 70
• D. 256

Answer 1

Answer: (B)

Let $A = \left(^{8}C_{1} - ^{8}C_{2} + ^{8}C_{3} -^{8}C_{4} + ^{8}C_{5} - ^{8}C_{6} +^{8}C_{7} - ^{8}C_{8}\right) $
$ = \frac{8!}{1!7! } - \frac{8!}{2!6!} + \frac{8!}{3!5!} - \frac{8!}{4!4!} + \frac{8!}{5!3!} - \frac{8!}{6!2!} + \frac{8!}{7!1!} - \frac{8!}{0!8!} $
Note : $^{n}C_{r} = \frac{n!}{r!\left(n-r\right)!} $
Thus, $A = 8 - \frac{8\times7}{2} + \frac{8\times 7\times 6}{3\times 2} - \frac{8\times 7\times 6\times 5}{4\times 3\times 2\times 1} + \frac{8\times 7\times 6}{3\times2}- \frac{8\times 7}{2} + 8 - 1 $
And A = 8 - 28 + 56 - 70 + 56 - 28 + 8 - 1 = 1