Navjyot Kalra
Last Activity: 10 Years ago
Hello Student,
Please find the answer to your question
Number of ways in which a student can select at least one and almost n books out of (2n + 1) books is equal to
= 2n + 1C1 + 2n + 1C2 + 2n + 1C3 + . . . . . . . . . . . . . . . . . . . + 2n + 1Cn
= 1/2 [2. 2n + 1C1 + 2. 2n + 1C2 + 2. 2n + 1C3 + . . . . . . . . . . . . . . . . . . . . . . + 2. 2n + 1Cn]
= 1/2 [(2n + 1C1 + 2n + 1C2n) + (2n + 1C2 + 2n + 1C2n - 1) + (2n + 1C3 + 2n + 1C2n – 2) + . . . . . . . . . . . . + (2n + 1Cn + 2n + 1Cn + 1)]
[Using n Cr = n Cn – r]
= 1/2 [2n + 1C1 + 2n + 1C2 2n + 1C3 + . . . . . . . . . . . . . + 2n + 1Cn + 2n + 1Cn + 1 + 2n + 1Cn + 2 + . . . . . . . . . . . . . . + 2n + 1C2n]
= 1/2 [2n + 1C0 + 2n + 1C1 + 2n + 1C2 + . . . . . . . . . . . . . . . . + 2n + 1C2n + 1 – 1 – 1]
= 1/2 [2 2n + 1 – 2] = 22n – 1
ATQ, 22n – 1 = 63 ⇒ 22n = 64 = 26
⇒ 2n = 6 ⇒ n = 3
Thanks
navjot kalra
askIITians Faculty
