Question

The value of
¹⁶C₉ + ¹⁶C₁₀- ¹⁶C₆ - ¹⁶C₇ is

• A. 0
• B. 1
• C. ¹⁷C₁₀
• D. ¹⁷C₃

Answer 1

The Correct Option is A

The value of \(^{16}C_9 + ^{16}C_{10} - ^{16}C_6 - ^{16}C_7\) is

We use the property \(^nC_r = ^nC_{n-r}\).

• \(^{16}C_9 = ^{16}C_7\)
• \(^{16}C_{10} = ^{16}C_6\)

Therefore, \(^{16}C_9 + ^{16}C_{10} - ^{16}C_6 - ^{16}C_7 = ^{16}C_7 + ^{16}C_6 - ^{16}C_6 - ^{16}C_7 = 0\)

 

Alternatively, using Pascal's identity \(^nC_r + ^nC_{r-1} = ^{n+1}C_r\):

\(^{16}C_9 + ^{16}C_{10} = ^{17}C_{10}\)

\(^{16}C_6 + ^{16}C_7 = ^{17}C_7\)

So the expression becomes \(^{17}C_{10} - ^{17}C_7 = ^{17}C_7 - ^{17}C_7 = 0\)

Answer: (A) 0

Answer 2

Using the property $ { }^{n}C_k = { }^{n}C_{n-k} $, we rewrite:

$$ { }^{16}C_9 + { }^{16}C_{10} - { }^{16}C_6 - { }^{16}C_7 = { }^{16}C_7 + { }^{16}C_6 - { }^{16}C_6 - { }^{16}C_7. $$

Simplify:

$$ { }^{16}C_7 + { }^{16}C_6 - { }^{16}C_6 - { }^{16}C_7 = 0. $$