Question:medium

If $^{n}C_{12} = ^{n}C_{8}$, then $^{n}C_{17} =$

Show Hint

Always simplify combinations using $^{n}C_r = ^{n}C_{n-r}$ before computing to minimize arithmetic operations ($^{20}C_{17} \to ^{20}C_3$).
Updated On: Jun 3, 2026
  • 1140
  • 1150
  • 1160
  • 1170
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: The combination matching rule.
If $^{n}C_x = {}^{n}C_y$ and $x \ne y$, then the two lower numbers must add up to $n$, that is $x + y = n$.

Step 2: Find $n$.
Here $x = 12$ and $y = 8$, which are different, so \[ n = 12 + 8 = 20 \]

Step 3: Rewrite the target.
We need $^{20}C_{17}$. Using $^{n}C_r = {}^{n}C_{n-r}$, this equals $^{20}C_3$, which is easier.

Step 4: Write the formula.
\[ ^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} \]

Step 5: Cancel before multiplying.
$18 \div 6 = 3$, so the top becomes $20 \times 19 \times 3$.

Step 6: Multiply.
\[ 20 \times 19 \times 3 = 1140 \] \[ \boxed{ ^{n}C_{17} = 1140 } \]
Was this answer helpful?
0

Top Questions on permutations and combinations