In how many ways can the letters of MANAGEMENT be arranged?
10 letters: M, A, N, E each twice
10! ÷ (2!)⁴ = 36,28,800 ÷ 16 = 2,26,800
Trap: 4,53,600 divides for only three pairs.
Answer: 2,26,800
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.
In how many ways can the letters of MANAGEMENT be arranged?
10 letters: M, A, N, E each twice
10! ÷ (2!)⁴ = 36,28,800 ÷ 16 = 2,26,800
Trap: 4,53,600 divides for only three pairs.
Answer: 2,26,800
How many arrangements can be made of the letters of SUCCESS?
7 letters: S × 3, C × 2
7! ÷ (3! × 2!) = 420
Trap: 840 divides only for the Ss.
Answer: 420
In how many arrangements of SUCCESS are the three Ss together?
SSS as one unit + U, C, C, E → 5 units with C twice
5!/2! = 60
Trap: 120 forgets the repeated C.
Answer: 60
A committee of 4 is chosen from 4 doctors and 6 engineers. In how many ways can it include at least one doctor?
All: 10C4 = 210
No doctor: 6C4 = 15
210 − 15 = 195
Answer: 195
A committee of 6 is chosen from 7 men and 4 women so that men outnumber women. In how many ways?
4M 2W: 35 × 6 = 210
5M 1W: 21 × 4 = 84
6M: 7
Total = 301
Trap: 210 takes only 4 men.
Answer: 301
How many 3-digit odd numbers have all their digits different?
Last digit odd: 5 ways
First digit: not 0 and not the last → 8
Middle: 8 left
5 × 8 × 8 = 320
Trap: 360 lets 0 lead.
Answer: 320
How many 3-digit numbers have digits that add up to 5?
a + b + c = 5 with a ≥ 1 → a′ + b + c = 4
(4 + 2)C2 = 15
Trap: 21 lets the first digit be 0.
Answer: 15
Signals are made by hoisting 1, 2 or 3 of 6 different flags one above another. How many signals are possible?
1 flag: 6; 2 flags: 6 × 5 = 30; 3 flags: 6 × 5 × 4 = 120
Total = 156
Trap: 41 counts selections, not orders.
Answer: 156
Seven-digit phone numbers must begin with 9 and have no repeated digit. How many such numbers are there?
First digit fixed as 9
Remaining 6 places from the other 9 digits: 9P6 = 60,480
Trap: 5,31,441 = 9⁶ allows repeats.
Answer: 60,480
A test has 8 multiple-choice questions with 4 options each. In how many ways can a student answer every question?
4 choices for each: 4⁸ = 65,536
Trap: 32 adds 8 × 4.
Answer: 65,536
Timed online tests with instant scores and explanations.