🧮 Aptitude & Reasoning · Chapter 7

Permutations, combinations, probability

Pick two players from fifteen and you get 105 pairs; name a captain and a vice-captain and the same fifteen give 210 — one question, does order matter, decides half the marks in this topic.

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The basics

Counting is finding how many ways something can happen without listing them; probability is that count divided by the count of everything that could happen.

The problem it solves: answering "how many teams, seatings, passwords?" and "what are the chances?" when the list would be thousands long.

The mental model: every counting question is a sequence of choices. If you make one choice and then another, multiply the options; if it's one or the other, add. Then ask one question: does order matter? If swapping two chosen things gives a new outcome (captain vs vice-captain, a PIN), it's a permutation. If it gives the same outcome (a committee, a hand of cards), it's a combination — divide out the orderings.

Probability sits on top: favourable ÷ total, with both counted the same way.

Explained

The selectors have 15 players.

Pick 2 for a net session: 15 × 14 / 2 = 105 pairs. Pick a captain and a vice-captain: 15 × 14 = 210.

Same two people, same 15 players — but Rohit-captain-Virat-vice is a different outcome from Virat-captain-Rohit-vice, while the net-session pair is the same pair either way. The order question doubled the answer. Ask it first, every time.

Four friends, one Shatabdi bay

Four college friends have seats 21–24 in a Shatabdi: two window seats, two aisle seats. Everyone wants a window.

Who gets the windows? Choose 2 of 4 — order doesn't matter yet: 6 ways. Which window is whose? The two winners swap or don't: ×2. The aisle seats? The other two swap or don't: ×2.

6 × 2 × 2 = 24 — exactly 4! = 4 × 3 × 2 × 1, the number of ways to seat four people in four seats.

That's the whole topic in one argument: a choice then a choice multiplies, and a combination (6) times the arrangements inside it (2 × 2) is a permutation.

A counting tree: three coin tosses

             toss 1    toss 2    toss 3     outcome
                      ┌── H ──┬── H        HHH ✓
             ┌── H ───┤       └── T        HHT ✓
             │        └── T ──┬── H        HTH ✓
   start ────┤                └── T        HTT ✓
             │        ┌── H ──┬── H        THH ✓
             └── T ───┤       └── T        THT ✓
                      └── T ──┬── H        TTH ✓
                              └── T        TTT ✗

   2 × 2 × 2 = 8 leaves (multiply at every fork)
   at least one H  = 7  →  7/8   = 1 − P(TTT) = 1 − 1/8
   exactly two H   = 3  →  3/8   = 3C2 / 8

The tree is the multiplication rule drawn out. You will never draw it in the exam — but every formula is a shortcut for counting its leaves.

The formulas, and when each applies

Counting

AND (then)  → multiply       OR (either) → add
nPr = n!/(n−r)!              order matters    5P3 = 60
nCr = n!/(r!(n−r)!)          order doesn't    5C3 = 10
nCr = nPr / r!               nCr = nC(n−r)    10C7 = 10C3
repeats: n!/(p! q! …)        BANANA = 6!/(3!2!) = 60
round table: (n−1)!          6 people → 120, not 720

Restrictions — standard method, then the shortcut

• Must be together: glue them into one block, arrange, then × arrangements inside the block
• Never together: total − together works for a pair; for "no two together" use the gap method — arrange the others, then drop the restricted ones into the gaps
• At least k: add the exact cases, or total − the cases below k (shortcut, fewer terms)

Probability

P = favourable / total          P(not A) = 1 − P(A)
independent AND → multiply      exclusive OR → add
P(A or B) = P(A) + P(B) − P(A and B)
without replacement: numerator AND denominator shrink

Solving steps

Most wrong answers come from step 2 or from counting favourable ordered but total unordered.

Three test questions, two minutes

Q1 (30 s). In how many arrangements of ORANGE are the vowels together? Glue O, A, E into one block: 4 units (block, R, N, G) → 4! = 24. Inside the block, 3! = 6. Total 144. As a probability: 144/720 = 1/5.

Q2 (45 s). A 5-member committee from 6 men and 4 women, at least 2 women. Standard: add exact cases 2W, 3W, 4W. Shortcut: total − (0 women) − (1 woman) = 252 − 6 − 60 = 186. Trap: "pick 2 women first, then any 3": 4C2 × 8C3 = 336 — it counts the same committee several times.

Q3 (25 s). Two dice. P(at least one six)? 1 − P(no six) = 1 − 25/36 = 11/36. Adding 1/6 + 1/6 = 12/36 double-counts (6,6).

Two aces in a row

A deck of 52 cards is shuffled. You draw two cards, one after the other, without putting the first back.

Traps examiners set

• P vs C confusion — "select" and "committee" are C; "arrange", "rank", "captain and vice" are P
• Overcounting "at least" — picking the minimum first, then "any" for the rest, counts one group many times (336 instead of 186). Use the complement
• Round table — fix one person; (n − 1)!, not n!
• Repeated letters — divide by the factorial of each repeat
• Mixing ordered and unordered — favourable counted as ordered pairs, total as unordered (or the reverse) gives nonsense
• With vs without replacement — "drawn together" means without
• Adding probabilities that overlap — king OR heart is 4 + 13 − 1 = 16 cards, not 17

Words to know

In two minutes

• A choice followed by another choice multiplies; a choice between alternatives adds.
• If swapping two chosen items gives a new outcome, it's a permutation (nPr); if not, a combination (nCr = nPr / r!).
• Repeated letters divide by the factorial of each repeat; round tables fix one seat, giving (n − 1)!.
• Handle restrictions first: glue items that must be together, use gaps for items that must be apart.
• 'At least' is fastest as total minus the cases that fail; picking the minimum first overcounts.
• Probability = favourable / total, both counted the same way; P(at least one) = 1 − P(none).
• Without replacement, the second draw depends on the first: 4/52 × 3/51, not (1/13)².

Test yourself

There are 10 questions, 3 puzzles and 5 recall cards for this lesson in the app — the recall cards come back on later days, just before you would forget.