Match the entries in Column-I representing inequalities in n with their values given in Column-II.
Column-I | Column-II |
(i) 16Cn + 16Cn+1 + 17Cn + 2 ≥ 18C2n–1 | [A] 15 |
(ii) 16Cn+5 ≤ 17Cn+6 | [B] 6 |
(iii) 12 × (nC6)2 ≤ 7 × (n+1C5) × (n+1C7) | [C] 7 |
(iv) 2 × (n–1C4 – n–1C3) < 5 × (n–2C2) | [D] 12 |
Text Solution
Verified by Experts(i) [B]; (ii) [B]; (iii) [A]; (iv) [B]
Ans.
(i) [B], [C]
(ii) [B], [C]
(iii) [A], [B], [C], [D]
(iv) [B], [C]
Sol. (i) 16 C n + 16 C n+1 + 17 C n+2 ≥ 18 C 2n–1
⇒ 18 C n+2 ≥ 18 C 2n–1
⇒ n + 2 ≥ 2n – 1 if n + 2 ≤ 9 and n + 2 ≤ 2n –1 if n + 2 ≥ 9
Hence n ≤ 3 if n ≤ 7 and n ≥ 3 if n ≥ 7
Also n + 2 ≤ 18, 2n –1 ≤ 18 ⇒ n ≤ 9
Hence possible values of n are 1, 2, 3, 6, 7, 8, 9
(ii) 16 C n+5 ≤ 17 C n+6
As r n C r = n. n–1 C r–1
∴ 17 C n+6 = 
16 C n+5
Hence 16 C n+5 ≤ 
16 C n+5
And n + 6 ≤ 17 ⇒ n ≤ 11
Hence n = 0, 1, 2, …., 11
(iii) 12 × ( n C 6 ) 2 ≤ 7 × ( n+1 C 5 ) × ( n+1 C 7 )
12 × 
≤ 7 × 
× 
2 ≤ 
× 
∴ 2(n –4) (n –5) ≤ (n + 1) 2 ∴ 2n
2 –18n + 40 – n 2 –2n – 1 ≤ 0
⇒ 3 ≤ n ≤ 17
(iv) 2 
< 5 × 
⇒ 2 
< 
⇒ 
< 
⇒ (n –1)(n – 8) < 30 ⇒ –n 2 –9n – 22 < 0
⇒ 2 < n < 11 but n – 1 ≥ 4 ∴ n ≥ 5
Hence 5 ≤ n < 11
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