MAH-CET 2026 — Mathematics PYQ

Step 1: Simplify the general term using the relationship between Permutations and Combinations

The mathematical relationship between a permutation (${}^n{P}_r$) and a combination (${}^n{C}_r$) is given by:

$${}^n{P}_r=r!\cdot \left({}^n{C}_r\right)$$

Rearranging this formula helps substitute the numerator of our given expression:

$$\frac{{}^n{P}_r}{r!}{=}^n{C}_r$$

Step 2: Rewrite the summation

Now, substitute this identity back into the original summation series:

$$\sum _{r=1}^n\frac{{}^n{P}_r}{r!}=\sum _{r=1}^n{}^n{C}_r$$

Expanding this summation term-by-term from $r=1$ to $r=n$:

$$\sum _{r=1}^n{}^n{C}_r{=}^n{C}_1{+}^n{C}_2{+}^n{C}_3+\cdots {+}^n{C}_n$$

Step 3: Apply the standard Binomial Coefficient identity

From the Binomial Theorem, we know that the sum of all binomial coefficients for a given exponent $n$ equals $2^n$:

$${}^n{C}_0{+}^n{C}_1{+}^n{C}_2+\cdots {+}^n{C}_n=2^n$$

Since our summation starts from $r=1$ instead of $r=0$, we need to isolate the required series by moving ${}^n{C}_0$ to the right-hand side:

$${}^n{C}_1{+}^n{C}_2+\cdots {+}^n{C}_n=2^n-{}^n{C}_0$$

We know that ${}^n{C}_0=1$. Substituting this value gives:

$$\sum _{r=1}^n{}^n{C}_r=2^n-1$$

Correct Answer

(b) $2^n-1$