Permutations & CombinationsJEE Main

Counting ordered and unordered selections — interactive Mathematics simulation for IIT-JEE.

Concept

Counting problems split into two kinds. When order matters you count permutations ( $^nP_r$ ); when it doesn't you count combinations ( $^nC_r$ ). Since each combination can be ordered in $r!$ ways, $^nP_r = {}^nC_r \cdot r!$ .

Key formula

^nP_r = \frac{n!}{(n-r)!}, \qquad ^nC_r = \frac{n!}{r!\,(n-r)!}

Derivation

Filling $r$ ordered slots from $n$ items gives $n(n-1)\cdots(n-r+1) = n!/(n-r)!$ choices — that is $^nP_r$ .

If the order of the chosen $r$ is irrelevant, every group is counted $r!$ times, so we divide: $^nC_r = {}^nP_r/r! = n!/[r!(n-r)!]$ . These combinations are exactly the entries of Pascal's triangle.

Scenarios to explore

Permutations & Combinations — nPr, nCr and Pascal's triangle.

Real-world applications

Probability (counting favourable vs total outcomes).
Binomial theorem coefficients.
Team selection, seating and arrangement problems.

JEE exam tips

$^nC_r = {}^nC_{n-r}$ — choose the smaller $r$ to compute faster.
Pascal's rule: $^nC_r = {}^{n-1}C_{r-1} + {}^{n-1}C_r$ .
Sum of a Pascal row $\sum_r {}^nC_r = 2^n$ .

Common mistakes

Using a permutation when order doesn't matter (over-counts by $r!$ ).
Forgetting $^nC_r = {}^nC_{n-r}$ (the symmetry of Pascal's triangle).
Treating $0! = 0$ — it is defined as $1$ .

Exam traps to avoid

'At least one' problems are often easier via the complement ( $2^n$ total − none).
Circular arrangements of $n$ objects give $(n-1)!$ , not $n!$ .