GMAT combinatorics strategies

A permutation counts arrangements where order matters: $P(n, r) = \frac{n!}{(n-r)!}$. A combination counts selections where order does not matter: $C(n, r) = \frac{n!}{r!(n-r)!}$. The only difference is the $r!$ in the denominator of the combination formula. That factor removes the duplicate arrangements that come from reordering the same selected items. Choosing 3 people from 10 is $C(10, 3) = \frac{10!}{3! \times 7!} = 120$. Arranging 3 people from 10 in a row is $P(10, 3) = \frac{10!}{7!} = 720$.

When a problem asks how many ways to arrange items, draw one slot for each position and fill in the number of choices per slot. Arranging 5 books on a shelf: 5 choices for the first slot, 4 for the second, 3 for the third, 2 for the fourth, 1 for the last. Multiply: $5 \times 4 \times 3 \times 2 \times 1 = 120$. The slot method is the same as the permutation formula but more intuitive because you see the choices shrink. For arrangements with restrictions (a specific person must be first), fix the restricted slot first, then fill the rest.

A club has 7 members. In how many ways can a president, vice president, and treasurer be chosen if no member can hold more than one position? Step 1: Does order matter? Yes. The president is different from the vice president, so the same three people in different roles counts as a different outcome. This is a permutation. Step 2: Identify $n$ and $r$. The pool is 7 members, and we choose 3 for the positions. $n = 7$, $r = 3$. Step 3: Apply the slot method. The president slot has 7 choices. The vice president slot has 6 (one member is already president). The treasurer slot has 5. Multiply: $7 \times 6 \times 5 = 210$. Step 4: Check for restrictions. No member holds more than one position, which the shrinking slots already enforce. The answer is 210.