If ⁿPᵣ = ⁿPᵣ₊₁ and ⁿCᵣ = ⁿCᵣ₋₁, then the value of r is equal to :

Solution

Maths sir · 4 steps

Each equation gives you one link between n and r, so solve the two together.

1. Permutation condition. n!/(n-r)! = n!/(n-r-1)! forces (n-r)! = (n-r-1)!. Since (n-r)! = (n-r)×(n-r-1)!, this means n - r = 1, so n = r + 1.
2. Combination condition. ⁿCₐ = ⁿC_b with a ≠ b happens only when a + b = n. Here r + (r-1) = n, so n = 2r - 1.

   r + 1 = 2r - 1 ⇒ r = 2, n = 3

3. Check. ³P₂ = 6 = ³P₃ and ³C₂ = 3 = ³C₁. Both conditions hold.
4. So. r = 2, which is option (c).

The idea to keep. Equal permutations means the leftover factorials match, so n - r = 1. Equal combinations ⁿCₐ = ⁿC_b means a + b = n. Turn each condition into a linear equation and solve.

Answer from the NTA final answer key. Solution by PrepWiser.