8.3 Probability, Permutations & Combinations

Probability and Counting on the CAE Academic Paper

FSc Pre-Engineering probability and combinatorics appear as rapid “how many ways” and “what is the probability” stems. The two skills share one exam skill: count carefully, then divide favorable by total when outcomes are equally likely.

Classical Probability

If a sample space S has n(S) equally likely outcomes and event E has n(E) favorable outcomes,

[P(E) = \frac{n(E)}{n(S)}.]

Worked die. Fair six-sided die: P(even) = 3/6 = 1/2. P(not 6) = 1 − 1/6 = 5/6.

Worked cards (no replacement thinking). From a standard 52-card deck, P(ace) = 4/52 = 1/13. P(heart or ace) = P(heart) + P(ace) − P(ace of hearts) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13.

Worked independent events. Two fair coins: P(both heads) = (1/2)·(1/2) = 1/4. Sample space {HH, HT, TH, TT} confirms one of four outcomes.

Conditional probability (light touch). P(A|B) = P(A ∩ B)/P(B) when P(B) > 0. If a die shows an even number, P(it is 2 | even) = (1/6)/(1/2) = 1/3 (favorable among {2,4,6}).

Worked bag without replacement. An urn holds 3 white and 2 black balls. Two balls are drawn in succession without replacement. P(first white and second black) = (3/5)·(2/4) = 3/10. P(both white) = (3/5)·(2/4) = 3/10 as well—same arithmetic, different event. P(same color) = P(WW) + P(BB) = (3/5)(2/4) + (2/5)(1/4) = 6/20 + 2/20 = 8/20 = 2/5.

Mutually exclusive vs independent (trap). Mutually exclusive events cannot both occur (A ∩ B = ∅), so for nontrivial events they are not independent. Independent events can both occur; their joint probability factors. If a stem says “cannot happen together,” use addition without the intersection term—do not multiply.

Factorials

n! = n·(n−1)·…·1, with 0! = 1 by convention. You need this for both permutations and combinations.

Permutations: Order Matters

The number of ways to arrange r distinct objects chosen from n distinct objects is

[P(n, r) = \frac{n!}{(n-r)!} = n(n-1)\cdots(n-r+1).]

Full arrangements of n distinct objects: P(n, n) = n!.

Worked permutations. How many 3-letter codes from {A,B,C,D,E} with no repetition? P(5, 3) = 5·4·3 = 60.

Worked ranking. 8 cadets; how many ways to award gold, silver, bronze (distinct medals)? P(8, 3) = 8·7·6 = 336.

With identical letters (arrangement formula). Distinct permutations of the letters in “PAF”: 3! = 6. For a word with repeated letters, divide by factorials of repeats—e.g. arrangements of “BOOK” = 4!/(2!) = 12. Arrangements of “SUCCESS” = 7!/(3!·2!) = 420 (S three times, C twice).

Circular permutations (occasional). Distinct seats around a round table for n people: (n − 1)! (rotations counted as the same). For 5 officers at a round briefing table: 4! = 24.

Combinations: Order Does Not Matter

The number of ways to choose r objects from n when order is irrelevant is

[C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!} = \frac{P(n,r)}{r!}.]

Symmetry: C(n, r) = C(n, n − r). Pascal: C(n, r) = C(n−1, r−1) + C(n−1, r).

Worked combinations. Choose 3 subjects from 7 electives: C(7, 3) = 35. Same as C(7, 4) = 35 by symmetry.

Worked committee. From 5 pilots and 4 engineers, form a team of 3 pilots and 2 engineers: C(5, 3)·C(4, 2) = 10·6 = 60.

Permutation vs Combination Decision Tree

Ask: Does rearranging the same people/objects create a new outcome?

• Yes (passwords, podium places, seating in a row) → permutation
• No (committee membership, lottery numbers as a set, choosing exam topics) → combination

Side-by-side. Selecting 2 officers from 5 for a patrol where roles are identical: C(5, 2) = 10. Selecting a patrol leader and a deputy from 5: P(5, 2) = 20.

Probability with Combinations

Worked hand probability. A box has 6 red and 4 blue bolts. Two bolts drawn at random without replacement. P(both red) = C(6, 2)/C(10, 2) = 15/45 = 1/3. Sequentially: (6/10)·(5/9) = 30/90 = 1/3—same answer.

Worked mixed. P(one red and one blue) = [C(6,1)C(4,1)]/C(10,2) = 24/45 = 8/15.

Binomial probability (quick bridge). If each independent trial has success probability p, then P(exactly k successes in n trials) = C(n, k) pᵏ (1−p)ⁿ⁻ᵏ. Example: fair coin, P(exactly 3 heads in 5 tosses) = C(5, 3)(1/2)⁵ = 10/32 = 5/16. Even if the full binomial theorem is not named on the slip, the C(n, k) factor is the same counting tool.

Worked with-replacement contrast. 4-digit PINs using digits 0–9 allowing repeats: 10⁴ = 10000. Without repeats: P(10, 4) = 10·9·8·7 = 5040. Wrong options often swap these two.

On the Exam

Expect: “P(n,r) = ?”, “C(n,r) = ?”, “probability both are …”, and “how many committees.” Compute small factorials by expanding products rather than writing n! fully. If stuck between 60 and 10 on a “choose 3 from 5” stem, re-read whether roles are distinct—that single word usually decides P vs C. Prefer complement counting when “at least one” appears: P(at least one) = 1 − P(none).