At a party every guest shook hands once with every other guest, and there were 66 handshakes. How many guests were there?
nC2 = 66 → n(n − 1) = 132 = 12 × 11
n = 12
Answer: 12
Permutations and Combinations · Exercise 2 of 5
40 medium permutations and combinations questions for bank, railway and placement tests: vowels apart or in fixed places, committees with conditions, collinear points, parallelograms, chessboards, stars and bars, alternate seating, a city grid and a grid of squares.
At a party every guest shook hands once with every other guest, and there were 66 handshakes. How many guests were there?
nC2 = 66 → n(n − 1) = 132 = 12 × 11
n = 12
Answer: 12
For a photo, 5 boys and 3 girls stand in a row. In how many ways can they stand if the three girls are together?
Girls as one block + 5 boys → 6! = 720
Girls inside the block: 3! = 6
720 × 6 = 4,320
Answer: 4,320
In how many ways can 4 boys and 4 girls stand in a row so that boys and girls alternate?
Patterns: BGBG… or GBGB… → 2
Boys 4! × girls 4! = 576 per pattern
2 × 576 = 1,152
Trap: 576 uses only one pattern.
Answer: 1,152
4 boys and 4 girls sit round a table, boys and girls alternating. In how many ways can they sit?
Seat the boys in a circle: 3! = 6
Girls in the 4 gaps: 4! = 24
6 × 24 = 144
Trap: 1,152 treats the table like a row.
Answer: 144
A basket has 3 apples, 4 bananas and 5 mangoes; fruits of one kind are identical. In how many ways can at least one fruit be chosen?
Apples: 0–3 (4 ways), bananas 5 ways, mangoes 6 ways
4 × 5 × 6 = 120, minus choosing nothing
= 119
Trap: 120 includes the empty choice.
Answer: 119
How many positive divisors does 360 have?
360 = 2³ × 3² × 5
(3 + 1)(2 + 1)(1 + 1) = 24
Answer: 24
In how many ways can one or more books be chosen from 10 different books?
Each book is in or out: 2¹⁰ = 1,024
Remove choosing none: 1,023
Trap: 1,024 includes the empty choice.
Answer: 1,023
3 different maths books and 2 different physics books are put on a shelf so that books of each subject stay together. In how many ways?
Order of subjects: 2!
Maths among themselves 3!, physics 2!
2 × 6 × 2 = 24
Trap: 12 forgets that the subjects can swap.
Answer: 24
8 people are to be seated at two different round tables of 4 seats each. In how many ways can this be done?
Choose 4 for the first table: 8C4 = 70
Each table: 3! circular arrangements
70 × 6 × 6 = 2,520
Trap: 1,260 treats the tables as identical.
Answer: 2,520
In how many ways can 10 players be split into two teams of 5 (the teams have no names)?
10C5 = 252 picks one team; the other is fixed
Each split is counted twice → 126
Trap: 252 counts each split twice.
Answer: 126
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