Counting problems on the GMAT are not about memorizing complex factorial formulas. They are about making a series of logical decisions: Does order matter? Are there constraints? This guide will teach you the 'Slot Method' to solve even the hardest P&C questions intuitively.
The most critical step is the first one. Learn to distinguish between a 'Selection' (Combination) and an 'Arrangement' (Permutation) using our visual Team vs. Photo scenarios.
The first step in any counting problem is to ask yourself one question: Does the order matter?
If I swap the positions of two items, do I get a new outcome or the same outcome?
You have 5 players (A, B, C, D, E). You need to select a team of 3 players.
Let's test the order:
Conclusion: Changing the order does NOT create a new team. Therefore, Order Does Not Matter.
This is a Combination problem.
Stop using nPr and nCr formulas blindly. Learn the intuitive 'Countdown Method' (Slot Method) to calculate arrangements and selections faster and with fewer errors.
Once you've identified the type of problem, how do you calculate the number? Forget the complex factorial formulas $$ (n! / r!(n-r)! )$$. We use the intuitive Countdown Method.
We want to select 3 players from a pool of 5 players.
In math terms: $$ n = 5, r = 3 $$
Start with the total available (5) and count down for the number of spots we need to fill (3 spots).
$$ 5 \times 4 \times 3 $$
Because order doesn't matter (A-B-C is the same as C-B-A), we have overcounted! We must divide to remove the duplicates.
Count down from the number of spots (3) to 1.
$$ 3 \times 2 \times 1 $$
$$ \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = \frac{60}{6} = 10 $$
Verbal Conclusion: There are exactly 10 distinct teams of 3 that can be formed from 5 players.
Real GMAT problems have rules. Learn how to handle scenarios where specific items MUST be chosen or MUST be excluded from the group.
Real-world problems often have rules. Some people must be on the team, others cannot be. Here is how to handle them.
Select a committee of 3 from 6 members (A, B, C, D, E, F) such that 'D' is never selected.
Since D cannot be on the team, we simply remove D from the pool of candidates entirely.
Original Pool: 6.
New Pool: 5 (A, B, C, E, F).
We now need to select 3 people from the remaining 5.
$$ ^5C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10 $$There are 10 such committees.
How do you seat people if two friends refuse to be separated? Learn the 'Glue' technique to treat multiple items as a single unit.
When items must stay together, treat them as a single unit.
Seat 8 friends (A through H) on 8 chairs such that C and H must sit together.
Imagine we glue C and H together into one 'Super-Person' [CH].
Now, instead of 8 people, we have:
A, B, D, E, F, G, [CH]
This is effectively 7 items to arrange.
Arranging 7 items takes $$ 7! $$ ways.
Don't forget inside the 'Super-Person'! C and H can swap places (CH or HC).
They can be arranged in $$ 2! $$ ways.
Multiply the external arrangement by the internal arrangement.
$$ \text{Total} = 7! \times 2! $$Verbal Conclusion: First arrange the groups, then arrange within the groups.
Calculating 'Never Together' directly is messy. Learn the powerful shortcut of subtracting the 'Together' cases from the Total to find the answer instantly.
Calculating 'never together' directly is hard. It is much easier to calculate 'Total' and subtract the 'Together' case.
Seat 5 friends (A, B, C, D, E) such that B and D are NEVER together.
Ignore the restriction. Arrange 5 people in 5 chairs.
$$ 5! = 120 \text{ ways} $$Calculate the opposite: How many ways do B and D sit TOGETHER?
Glue [BD]. Now we have {A, C, E, [BD]} = 4 items.
$$ 4! \times 2! = 24 \times 2 = 48 \text{ ways} $$$$ \text{Never Together} = \text{Total} - \text{Always Together} $$
$$ 120 - 48 = 72 $$There are 72 ways to seat them apart.
What happens when the items aren't unique (like the letters in MISSISSIPPI)? Learn how to adjust your calculation to account for duplicates.
When items are identical (like duplicate letters), swapping them doesn't create a new arrangement. We must divide to remove these invisible swaps.
Divide by the factorial of the repetition count.
$$ \frac{5!}{2!} $$In Data Sufficiency, your goal is not to find the number, but to determine if it CAN be found. Learn to spot the sufficiency without wasting time on massive calculations.
In DS, your goal is to determine IF a unique number can be found. You almost never need to actually multiply the factorials.
How many codes can be formed using digits 0-9?
(1) No digit can be repeated.
(2) The code is a 5-digit code and must be odd.
"No repeat" is a constraint. But do we know anything else? Can it start with 0? What is the length of the code?
INSUFFICIENT.
"Code must be odd". This affects the last digit. But can we repeat digits? We don't know.
Case A: Repeats allowed $$ 9 \times 10 \times 10 \times 10 \times 5 $$
Case B: No repeats $$ 8 \times 8 \times 7 \times 6 \times 5 $$
INSUFFICIENT.
We know: 5 digits, No repeats, Odd.
1. Last digit: 5 options (1,3,5,7,9).
2. First digit: 8 options (can't be 0, can't be last digit).
3. Middle 3 digits: Pick from remaining 8.
We have a specific number of choices for every slot. We CAN calculate a unique value.
SUFFICIENT. (Answer C)
Put your understanding of GMAT Permutations & Combinations: A Logic-Based Approach to the test with our premium, adaptive practice questions. Build pacing efficiency and eliminate knowledge gaps now.
Stop calculating sums! Learn the 'Average of First & Last' rule to solve Arithmetic Progression (AP) problems in seconds.
Recursive sequences look scary, but they follow a pattern. Learn how to decode the definition and spot the hidden loop.
Why feeling the OG is easy is actually a sign of success, and how to use the 'Easy' questions to fix your pacing.