Paper IPaper I · General Mental Ability

Permutation, Combination and Probability

Factorials, the fundamental counting principle, nPr and nCr, arrangements, selections, and basic probability, with worked examples and practice

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PaperPaper ISubjectGMASyllabusGeneral Mental Ability: quantitative aptitude including numerical abilityImportanceMedium
GMAQuantitative AptitudePermutationCombinationProbabilityCountingPaper 1

Counting problems and the chance built on them. The single decision that settles almost every question is: does order matter? If yes, it is a permutation (arrangement); if no, it is a combination (selection). Probability then divides favourable outcomes by total outcomes. This sits on the multiplication and factorial ideas in number system and simplification.

Factorials and the counting principle

The factorial n! is the product of all whole numbers from 1 to n. By convention 0! = 1 and 1! = 1.

n n!
0 1
1 1
2 2
3 6
4 24
5 120
6 720

Fundamental counting principle: if one task can be done in m ways and a second in n ways, the two together can be done in m times n ways. Add when the choices are alternatives ("or"); multiply when they happen in sequence ("and").

Permutation and combination

Concept Formula
Permutations of n distinct items taken r at a time nPr = n! / (n minus r)!
Combinations of n distinct items taken r at a time nCr = n! / [ r! times (n minus r)! ]
Link between them nPr = nCr times r!
Arrange all n distinct items n!
Arrange n items with repeats n! / (p! times q! ...) for repeated groups of size p, q, ...
Circular arrangement of n items (n minus 1)!
nC0 and nCn both equal 1
nCr equals nC(n minus r) symmetry property

Worked identities worth memorising

  • nC1 = n.
  • nC2 = n(n minus 1) / 2.
  • Sum of all subsets: nC0 + nC1 + ... + nCn = 2 to the power n.

Probability basics

Concept Rule
Probability of an event E P(E) = favourable outcomes / total outcomes
Range 0 (impossible) to 1 (certain)
Complement P(not E) = 1 minus P(E)
Mutually exclusive (cannot both happen) P(A or B) = P(A) + P(B)
Not mutually exclusive P(A or B) = P(A) + P(B) minus P(A and B)
Independent events P(A and B) = P(A) times P(B)

Standard sample spaces: a coin has 2 outcomes, a die has 6, a standard deck has 52 cards (26 red, 26 black, 4 suits of 13, 12 face cards counting J, Q, K).

Worked examples

Example 1: Arrangement (order matters)

In how many ways can 5 recruits stand in a row for a photo?

All 5 distinct items arranged: 5! = 120 ways.

Example 2: Selection (order does not matter)

From 8 officers, a committee of 3 is to be chosen. How many committees?

8C3 = 8! / (3! times 5!) = (8 times 7 times 6) / (3 times 2 times 1) = 336 / 6 = 56.

Example 3: Arrangement with repeated letters

How many distinct arrangements of the letters of the word LEVEL?

LEVEL has 5 letters: L appears twice, E appears twice, V once. Arrangements = 5! / (2! times 2!) = 120 / 4 = 30.

Example 4: Circular arrangement

In how many ways can 6 guests be seated around a round table?

Circular arrangement of 6 = (6 minus 1)! = 5! = 120.

Example 5: Probability with a die

A fair die is rolled. What is the probability of getting a prime number?

Primes on a die are 2, 3, 5, so 3 favourable outcomes out of 6. P = 3/6 = 1/2.

Example 6: Probability with cards

One card is drawn from a standard pack of 52. What is the probability it is a king or a heart?

Kings = 4, hearts = 13, but the king of hearts is counted in both, so subtract 1. P = 4/52 + 13/52 minus 1/52 = 16/52 = 4/13.

Example 7: Independent events

Two fair coins are tossed. What is the probability of two heads?

Each head has probability 1/2; independent, so P = 1/2 times 1/2 = 1/4.

Shortcut tips

  • First ask: does order matter? Race positions, passwords, and seat orders are permutations; teams, committees, and handshakes are combinations.
  • nCr = nC(n minus r), so to choose 8 from 10, compute 10C2 = 45 instead of the larger 10C8.
  • Handshake problems are nC2 selections: n people shaking hands give nC2 = n(n minus 1)/2 handshakes.
  • For "at least one" probability, it is usually faster to find P(none) and subtract from 1.
  • Keep probabilities as fractions until the last step to avoid rounding error, then convert if a decimal is asked.

Practice questions

  1. Evaluate 6P2.
  2. Evaluate 7C3.
  3. In how many ways can the letters of the word APPLE be arranged?
  4. From 10 players, how many ways to pick a team of 4?
  5. In how many ways can 4 people be seated at a round table?
  6. A die is rolled; find the probability of an even number.
  7. A bag has 4 red and 6 blue balls; one is drawn. Find the probability it is red.
  8. Two dice are thrown; find the probability the sum is 7.
  9. A card is drawn from a pack of 52; find the probability it is a face card.
  10. There are 12 people in a room; if everyone shakes hands once with everyone else, how many handshakes occur?

Answer key

Reveal the answer key and full worked solutions
  1. 6P2 = 6! / 4! = 6 times 5 = 30.
  2. 7C3 = (7 times 6 times 5) / (3 times 2 times 1) = 210 / 6 = 35.
  3. APPLE has 5 letters with P twice: 5! / 2! = 120 / 2 = 60.
  4. 10C4 = (10 times 9 times 8 times 7) / (4 times 3 times 2 times 1) = 5040 / 24 = 210.
  5. (4 minus 1)! = 3! = 6.
  6. Even numbers 2, 4, 6 give 3/6 = 1/2.
  7. 4 red out of 10 total = 4/10 = 2/5.
  8. Sums of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 outcomes out of 36 = 6/36 = 1/6.
  9. Face cards = 12 (J, Q, K in 4 suits) out of 52 = 12/52 = 3/13.
  10. 12C2 = 12 times 11 / 2 = 66 handshakes.

See also

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