Counting and probability questions appear in both Problem Solving and Data Sufficiency, and they punish formula-first thinking. The formulas themselves are simple: combinations count selections where order does not matter, permutations count arrangements where it does, and a probability is favourable outcomes divided by total outcomes when every outcome is equally likely. The hard part is deciding which situation you are in.
Ask two questions before you write anything. Does order matter? A committee is a selection, and so is a code whose digits must increase, even though it looks like an arrangement. Is there replacement? Drawing without replacement changes the total on every draw.
Two strategies crack most hard questions. The first is the complement: “at least one” or “at least two” is usually easier to count as the total minus the cases you do not want. The second is splitting the problem into cases that do not overlap and adding them. The classic trap is double-counting, for example picking “two women first, then anyone” for an at-least-two-women committee, which counts the same committee several times.
For probability, you can count unordered selections or multiply step-by-step probabilities, but stay consistent. Mixing ordered and unordered counts in one fraction is the most common way to land on a wrong answer that is waiting among the choices.
4 practice questions
Problem SolvingCombinationsMedium-hard
Question 1
A $4$-person committee is to be selected from a group of $6$ men and $5$ women. How many different committees include at least $2$ women?
A$215$
B$230$
C$265$
D$330$
E$360$
Show answer & explanation
Correct answer: A
Key Concept or Approach
Use the complement: all committees minus those with 0 or 1 woman.
Common trap: 360 - choosing 2 women first and then any 2 of the remaining 9 people, which double-counts committees with 3 or 4 women.
Problem SolvingCombinationsMedium-hard
Question 2
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 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).
Data SufficiencyProbabilityFractions/Ratios/DecimalsVery hard
Question 4
A bag contains only red chips and green chips, with at least one chip of each color. Two chips are drawn at random without replacement. Is the probability that both chips are red greater than $\frac{1}{2}$?
(1) The number of red chips is more than twice the number of green chips.
(2) There are fewer than 10 chips in the bag.
AStatement (1) ALONE is sufficient, but statement (2) ALONE is NOT sufficient.
BStatement (2) ALONE is sufficient, but statement (1) ALONE is NOT sufficient.
CBOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
DEACH statement ALONE is sufficient.
EStatements (1) and (2) TOGETHER are NOT sufficient.
Show answer & explanation
Correct answer: E
Key Concept or Approach
With $r$ red and $g$ green chips, $P(\text{both red}) = \frac{r(r-1)}{(r+g)(r+g-1)}$. Without replacement, small bags push this probability down, so test small cases.
Step-by-Step Solution
Step 1: Statement (1): $r = 3, g = 1$: $P = \frac{3 \cdot 2}{4 \cdot 3} = \frac{1}{2}$, not greater, so no. $r = 6, g = 1$: $P = \frac{30}{42} = \frac{5}{7}$, so yes. Not sufficient.
Step 2: Statement (2): the same two bags (4 chips and 7 chips) both have fewer than 10 chips, giving no and yes. Not sufficient.
Step 3: Together: both bags above satisfy both statements ($3 > 2$, $6 > 2$; 4 and 7 chips), yet one gives exactly $\frac{1}{2}$ and the other $\frac{5}{7}$. Not sufficient.
Final Answer
Answer: E
Common trap: C: students test only bags like 6 red and 1 green, where the probability is clearly above 1/2, and miss the 3-red/1-green bag, which gives exactly 1/2.
