Q100 / 180 · COMEDK 2022 · Mathematics
If ${ }^n C_3=220$, then $n=$ ?
- A11
- Correct answerB12
- C10
- D9
Solution
The formula for the combination is $^{n}C_{r} = \dfrac{n!}{r!(n-r)!}$.
Given $^{n}C_{3} = 220$, we have $\dfrac{n(n-1)(n-2)}{3 \times 2 \times 1} = 220$.
$n(n-1)(n-2) = 220 \times 6 = 1320$.
We look for three consecutive integers whose product is $1320$.
Testing values:
$10 \times 9 \times 8 = 720$
$11 \times 10 \times 9 = 990$
$12 \times 11 \times 10 = 1320$.
Therefore, $n = 12$.
Answer: 12
Given $^{n}C_{3} = 220$, we have $\dfrac{n(n-1)(n-2)}{3 \times 2 \times 1} = 220$.
$n(n-1)(n-2) = 220 \times 6 = 1320$.
We look for three consecutive integers whose product is $1320$.
Testing values:
$10 \times 9 \times 8 = 720$
$11 \times 10 \times 9 = 990$
$12 \times 11 \times 10 = 1320$.
Therefore, $n = 12$.
Answer: 12
Hence, the correct answer is option: 'B'.