A box holds 5 different red balls and 3 different white balls. In how many ways can 4 balls be chosen with at least one of each colour?
All: 8C4 = 70
All red: 5C4 = 5; all white impossible
70 − 5 = 65
Answer: 65
Permutations and Combinations · Exercise 4 of 5
65 moderate and difficult permutations and combinations questions, with points on a circle to read, in the style of IBPS / SBI PO and Clerk, RRB NTPC and Group D, UPSC CSAT, state PSC and IT campus placement tests.
A box holds 5 different red balls and 3 different white balls. In how many ways can 4 balls be chosen with at least one of each colour?
All: 8C4 = 70
All red: 5C4 = 5; all white impossible
70 − 5 = 65
Answer: 65
How many different bracelets can be made from 6 beads of different colours?
Circular: 5! = 120; a bracelet can be flipped → ÷ 2
= 60
Trap: 120 treats it like a seating plan.
Answer: 60
A host sits in a fixed chair at the head of a round table for 8. His 7 guests take the other seats, but the guest of honour must sit on the host's right. In how many ways?
Host's chair fixed, guest of honour fixed beside him
Other 6 guests in 6 seats: 6! = 720
Trap: 5,040 lets the guest of honour sit anywhere.
Answer: 720
4 different maths books and 3 different English books go on a shelf with all the English books together. In how many ways?
English block + 4 maths → 5! = 120
Inside the block: 3! = 6
120 × 6 = 720
Answer: 720
4 different maths books and 3 different English books are arranged so that no two English books are side by side. In how many ways?
Maths books: 4! = 24
5 gaps; English books in 3 of them, in order: 5P3 = 60
24 × 60 = 1,440
Answer: 1,440
A team of 3 is chosen from 4 boys and 3 girls with at least one girl. In how many ways?
All: 7C3 = 35
All boys: 4C3 = 4
35 − 4 = 31
Answer: 31
In how many ways can the letters of REGULATION be arranged so that the vowels occupy only the even positions?
10 letters, 5 vowels (E, U, A, I, O), 5 even places
Vowels: 5! = 120; consonants in odd places: 5! = 120
120 × 120 = 14,400
Answer: 14,400
How many arrangements of the letters of REGULATION begin with R and end with N?
Fix R and N; arrange the other 8: 8! = 40,320
Answer: 40,320
In how many ways can 3 different numbers be chosen from 1 to 20 so that their sum is even?
All three even: 10C3 = 120
Two odd, one even: 10C2 × 10 = 450
Total = 570
Answer: 570
In how many ways can 5 identical chocolates be shared among 3 children if some may get none?
Stars and bars: (5 + 2)C2 = 21
Trap: 6 = 4C2 gives each child at least one.
Answer: 21
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