Counting questions ask how many ways something can happen, and on the GMAT Focus they appear in both Quant Problem Solving and Data Insights Data Sufficiency. The arithmetic is light. The difficulty is deciding what exactly you are counting.
Start with one question: does order matter? Choosing a committee, a hand of cards or a set of toppings is a selection, so use combinations (nCr). Filling ranked positions, forming a code or seating people in a row is an arrangement, so use permutations or the slot method: write one slot per position and multiply the number of options for each. When some items are identical, divide by the factorial of each repeat count.
Restrictions change the method. If two people must sit together, glue them into one block and multiply by their internal arrangements. If two people must not sit together, count all arrangements and subtract the ones where they do. “At least one” almost always calls for the complement: total minus none.
The classic trap is double-counting. Picking “one woman first, then anyone else” for a committee that needs at least one woman counts the same group several times. When a count feels too large, list a small case by hand and compare.
4 practice questions
Problem SolvingCombinationsMedium-hard
Question 1
A lock code consists of $5$ different digits chosen from $1$ through $9$, arranged so that the digits increase from left to right. How many such codes contain the digit $5$?
A$56$
B$64$
C$70$
D$84$
E$126$
Show answer & explanation
Correct answer: C
Key Concept or Approach
Because the order is forced (increasing), each code is just a 5-element subset of $\{1,\dots,9\}$.
Step-by-Step Solution
Step 1: All codes: $\binom{9}{5} = 126$.
Step 2: Codes without 5: choose from the other 8 digits, $\binom{8}{5} = 56$.
Common trap: 56 - that counts the codes WITHOUT the digit 5.
Problem SolvingCombinationsMedium-hard
Question 2
A manager will assign $6$ different tasks to $3$ employees (Kim, Luis, and Maya) so that each employee receives exactly $2$ of the tasks. In how many different ways can the tasks be assigned?
A$15$
B$20$
C$45$
D$90$
E$729$
Show answer & explanation
Correct answer: D
Key Concept or Approach
The employees are distinct, so choose Kim's pair, then Luis's pair from what is left; Maya gets the rest. Do not divide by $3!$, because the groups are labeled.
Step-by-Step Solution
Step 1: Kim: $\binom{6}{2} = 15$ ways.
Step 2: Luis: $\binom{4}{2} = 6$ ways; Maya takes the last $2$ in $1$ way.
Step 3: Total: $15 \times 6 \times 1 = 90$.
Final Answer
Answer: D
Common trap: 15 - dividing 90 by 3! as if the three pairs were unlabeled, even though each pair goes to a specific person.
Problem SolvingProbabilityCombinationsHard
Question 3
A bag contains $4$ red, $3$ blue, and $2$ green marbles. If $3$ marbles are drawn at random without replacement, what is the probability that the three marbles are all different colors?
A$\frac{1}{6}$
B$\frac{16}{81}$
C$\frac{2}{9}$
D$\frac{2}{7}$
E$\frac{1}{3}$
Show answer & explanation
Correct answer: D
Key Concept or Approach
Count favorable unordered selections over all unordered selections.
Step-by-Step Solution
Step 1: Total selections: $\binom{9}{3} = 84$.
Step 2: One of each color: $4 \times 3 \times 2 = 24$.
Step 3: Probability $= \frac{24}{84} = \frac{2}{7}$.
Final Answer
Answer: D
Common trap: 16/81 - computing as if drawing with replacement (6 x 4 x 3 x 2 / 9^3).
Problem SolvingProbabilityCombinationsHard
Question 4
A box contains $6$ light bulbs, exactly $2$ of which are defective. An inspector tests the bulbs one at a time, in random order, without replacement. What is the probability that the second defective bulb is found on the fourth test?
A$\frac{1}{15}$
B$\frac{1}{6}$
C$\frac{1}{5}$
D$\frac{4}{15}$
E$\frac{1}{3}$
Show answer & explanation
Correct answer: C
Key Concept or Approach
A random testing order puts the two defective bulbs in a random pair of the $6$ positions; all $\binom{6}{2}$ pairs are equally likely.
Step-by-Step Solution
Step 1: Possible position pairs for the defective bulbs: $\binom{6}{2} = 15$.
Step 2: "Second defective on test 4" means one defective is in position $4$ and the other in position $1$, $2$, or $3$: $3$ pairs.
Step 3: Probability $= \frac{3}{15} = \frac{1}{5}$. Check sequentially: $P(\text{exactly one defective in tests 1-3}) \times P(\text{test 4 defective}) = \frac{\binom{2}{1}\binom{4}{2}}{\binom{6}{3}} \times \frac{1}{3} = \frac{12}{20} \times \frac{1}{3} = \frac{1}{5}$.
Final Answer
Answer: C
Common trap: 1/15 - counting only one favorable arrangement (e.g., defectives exactly at tests 3 and 4) instead of all three.
