A committee of 2 men and 3 women is to be chosen from 5 men and 6 women. In how many ways can this be done?
5C2 × 6C3 = 10 × 20 = 200
Trap: 462 is any 5 of the 11 people.
Answer: 200
Permutations and Combinations · Exercise 1 of 5
40 permutations and combinations questions at competitive-exam level: arranging letters with repeats, numbers from digits, circular seating and necklaces, committees, handshakes, diagonals, lines and triangles, and a street map – each solved one step per line with the common trap named.
A committee of 2 men and 3 women is to be chosen from 5 men and 6 women. In how many ways can this be done?
5C2 × 6C3 = 10 × 20 = 200
Trap: 462 is any 5 of the 11 people.
Answer: 200
A group of 4 is chosen from 6 men and 4 women. In how many ways can it have exactly 2 women?
4C2 × 6C2 = 6 × 15 = 90
Trap: 210 is any group of 4.
Answer: 90
A person has 5 shirts, 4 trousers and 3 ties. How many different outfits of one of each can be made?
Multiply the choices: 5 × 4 × 3 = 60
Trap: 12 adds them.
Answer: 60
There are 3 roads from A to B and 4 roads from B to C. In how many ways can one go from A to C and back to A without using any road twice?
Going: 3 × 4 = 12
Returning on unused roads: 3 × 2 = 6
12 × 6 = 72
Trap: 144 lets the same roads be reused.
Answer: 72
In how many ways can 3 boys and 2 girls stand in a row if the two girls must be together?
Girls as one block + 3 boys → 4! = 24
Girls can swap: × 2 = 48
Trap: 24 forgets the swap.
Answer: 48
In how many ways can 3 boys and 2 girls stand in a row if the girls are not next to each other?
All = 5! = 120
Girls together = 48
Not together = 120 − 48 = 72
Answer: 72
How many even 4-digit numbers can be formed with the digits 1, 2, 3, 4, 5 without repetition?
Last digit 2 or 4: 2 ways
Remaining places: 4 × 3 × 2 = 24
2 × 24 = 48
Trap: 120 ignores the even condition.
Answer: 48
How many 3-digit numbers have all three digits different?
First digit 1–9: 9 ways
Second: 9 (0 now allowed); third: 8
9 × 9 × 8 = 648
Trap: 720 lets 0 lead.
Answer: 648
From 5 vowels and 8 consonants, in how many ways can a set of 2 vowels and 3 consonants be selected?
5C2 × 8C3 = 10 × 56 = 560
Trap: 67,200 also arranges them into words.
Answer: 560
In how many ways can 4 different prizes be given to 3 students if a student may win any number of prizes?
Each prize goes to one of 3 students: 3⁴ = 81
Trap: 64 = 4³ reverses the roles.
Answer: 81
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