Option : (B)
(7C0 + 7C1) + (7C1 + 7C2) + (7C2 + 7C3) + (7C3 + 7C4) + (7C4 + 7C5) + (7C5 + 7C6) + (7C6 + 7C7)
= 1 + 2 x 7C1 + 2 x 7C2 + 2 x 7C3 + 2 x 7C4 + 2 x 7C5 + 2 x 7C6 + 1
= 1 + 2 x 7C1 + 2 x 7C2 + 2 x 7C3 + 2 x 7C3 + 2 x 7C2 + 2 x 7C6 + 1
⇒ (7C3 = 7C4 and 7C2 = 7C5)
= 2 + 22 (7C1 + 7C2 + 7C3)
= 2 + 22 (7 + \(\frac{7}{2}\) x 6 + \(\frac{7}{3}\) x \(\frac{6}{2}\) x 5)
= 2 + 252
= 254
= 28 – 2