Question

The least value of \( n \) such that \( {}^{n-1}C_6 + {}^{n-1}C_7<{}^nC_8 \) is

• A. \( 14 \)
• B. \( 15 \)
• C. \( 16 \)
• D. \( 17 \)

Hint:

Use the Pascal's identity \( {}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1} \) to simplify the inequality. Remember the property that binomial coefficients \( {}^nC_r \) increase with \( r \) up to the middle term(s) and then decrease. \( {}^nC_r<{}^nC_{r+1} \) when \( r<(n-1)/2 \).

Answer 1

The Correct Option is C

We use the identity \( {}^nC_r + {}^nC_{r+1} = {}^{n+1}C_{r+1} \).
Given the inequality \( {}^{n-1}C_6 + {}^{n-1}C_7<{}^nC_8 \).
Using the identity on the left side: $$ {}^{n-1}C_6 + {}^{n-1}C_7 = {}^nC_7 $$ So the inequality becomes \( {}^nC_7<{}^nC_8 \).
We know that \( {}^nC_r<{}^nC_{r+1} \) if \( r<\frac{n - 1}{2} \).
In our case, \( r = 7 \), so we need \( 7<\frac{n - 1}{2} \).
$$ 14<n - 1 $$ $$ n>15 $$ The least integer value of \( n \) satisfying this is \( n = 16 \).
Let's verify for \( n = 16 \): \( {}^{15}C_6 + {}^{15}C_7 = {}^{16}C_7 \) We need to check if \( {}^{16}C_7<{}^{16}C_8 \).
We know \( {}^nC_r<{}^nC_{r+1} \) if \( r<\frac{n}{2} \).
Here \( n = 16, r = 7 \).
Since \( 7<\frac{16}{2} = 8 \), we have \( {}^{16}C_7<{}^{16}C_8 \).
Let's check for \( n = 15 \): \( {}^{14}C_6 + {}^{14}C_7 = {}^{15}C_7 \) We need to check if \( {}^{15}C_7<{}^{15}C_8 \).
Here \( n = 15, r = 7 \).
Since \( 7 = \frac{15 - 1}{2} \), we have \( {}^nC_r = {}^nC_{r+1} \).
\( {}^{15}C_7 = {}^{15}C_8 \), so the inequality is not satisfied.
Thus, the least value of \( n \) is 16.