Probability & Permutations — PPSC previous year questions

Each handshake joins 2 of the 10 people, so count the number of pairs. > ¹⁰C₂ = (10 × 9) ÷ 2 = 45 There are 45 handshakes.

Read the 54% chance of passing both subjects against the 60% chance of passing English. Given that he passes English, the chance that he also passes Mathematics is the conditional probability. > P(Maths | English) = 0.54 ÷ 0.60 = 0.90 > P(fails Maths) = 1 − 0.90 = 0.10 = 10% So (d) 10.

Men cannot use the ladies' seat. If the woman takes the reserved seat, the two men are arranged in 2 of the other 4 seats: 4 × 3 = 12 ways. If she takes one of the 4 unreserved seats, the men are arranged in 2 of the remaining 3 unreserved seats: 4 × (3 × 2) = 24 ways. > woman on reserved seat: 4P2 = 12 > woman on other seat: 4 × 3P2 = 24 > total = 12 + 24 = 36.

Handshakes happen only between a man and a woman. So with m men and w women, m × w = 24. Hugs happen within each sex, so hugs = C(m,2) + C(w,2). Test every factor pair of 24. > 1 × 24 → 0 + 276 = 276 > 2 × 12 → 1 + 66 = 67 > 3 × 8 → 3 + 28 = 31 > 4 × 6 → 6 + 15 = 21 Of these values, only 21 appears among the choices.

Write every such number with five digit slots, using leading zeros — 47 becomes 00047. > Count d1 + d2 + d3 + d4 + d5 ≤ 6, each digit 0–9 > The total never reaches 10, so no digit can break its 0–9 limit; the count is C(6+5, 5) = C(11, 5) = 462 > Drop 00000, which is not a natural number There are 461 such numbers.

MIDDLE has 6 cards, but the two D cards look the same. > 6 cards in a line can go in 6! orders > Swapping the two identical Ds makes no new-looking line, so halve it The count is 6!/2.

A D belongs in the 3rd or the 4th place, so "one D placed correctly" can happen in 2 ways. > For each of them the other 5 cards fill the other 5 places in 5! = 120 ways > Exactly 1 of those 120 spells MIDDLE, so 119 spell it wrong > 2 × 119 There are 238 such placings.

A ticket is specific to both the starting station and the destination, so the journey A→B needs a different ticket from B→A. The number of ordered pairs of distinct stations is 20P2 = 20 × 19 = 380. (190 would be the answer if a ticket could be used in either direction.)

By the fundamental principle of counting, each of the 4 routes from A to B can be combined with any of the 6 routes from B to C. > Total ways = 4 × 6 = 24.

> Each coin is either used or not: 2⁴ = 16 ways > Remove the case where no coin is used: 16 − 1 = 15 Since all four denominations differ, every choice gives a different sum, so 15 sums are possible.