Q1 if 12C3=12Ck find k not equal to 3 Q 2 show that 7C1,7C2,7C3 are in AP - Maths - Permutations and Combinations

Question

Q1 if 12C3=12Ck find
k not equal to 3
Q 2 show that
7C1,7C2,7C3
are in AP

Ajanta Trivedi · Accepted Answer

(1)

Since ${}^{n}{C_{r} =} {}^{n}{C_{n - r}}$

therefore

${}^{12}{C_{3}} = {}^{12}{C_{k}} \Rightarrow k = 12 - 3 = 9$

(2)

${}^{n}{C_{r}} = \frac{n!}{r! (n - r)!} {}^{7}{C_{1}} = 7, {}^{7}{C_{2}} = \frac{7!}{2! (5)!} = \frac{7 \times 6}{2} = 21 {}^{7}{C_{3}} = \frac{7!}{3! (4)!} = \frac{7 \times 6 \times 5}{3 \times 2} = 35$

since the difference between 7, 21 and 21, 35 is same and i.e. 14.

therefore they are in AP.

Q1. if 12C3=12Ck find k not equal to 3 Q.2.show that 7C1,7C2,7C3 are in AP

Q1. if ¹²C₃=¹²C_kfind k not equal to 3

Q.2.show that ⁷C₁,⁷C₂,⁷C₃ are in AP