Practice Questions – Permutation and Combination – 1
1) Find the value of 9P3.
(a) 504
(b) 309
(c) 405
(d) 600
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(a) 504
Explanation:
9P3 = 9!/6!
Because, nPr
= (9 ×8 ×7 ×6!)/6!
= 9 × 8 × 7 = 504
2) Find the value of n, if nP5 = 20 nP3.
(a) 5
(b) 8
(c) 6
(d) 4
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(b) 8
Explanation:
nP5 = 20 nP3
(n – 3)(n – 4) = 20
n = 8 (n = -1 is inadmissible)
3) How many numbers greater than 5000 can be formed with the digits, 3, 5, 7, 8, 9 no digit being repeated?
(a) 216
(b) 126
(c) 512
(d) 252
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(a) 216
Explanation:
= 4 × 4 × 3 × 2 = 96
Thousands place cannot assume 3 since required numbers are greater than 5000.
= 5 × 4 × 3 × 2 × 1 = 120
Total required numbers = 96 + 120 = 216.
4) How many 6 digit telephone numbers can be constructed with the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each number starts with 35 and no digit appears more than once?
(a) 1600
(b) 1680
(c) 900000
(d) 9000
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(b) 1680
Explanation:
= 8 × 7 × 6 × 5 = 1680
3 and 5 have been already used, so we have 8 digits for thousands place, then 7 digits for hundreds place, 6 digits for tens digit and remaining 5 digits for unit place.
5) How many 6 digit numbers can be formed from the digits 1, 2, 4, 5, 6, 7 (no digit being repeated) which are divisible by 5?
(a) 555
(b) 156
(c) 120
(d) None of these
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(c) 120
Explanation:
A number is divisible by 5 only if its unit digit is either 5 or 0. So we fix 5 as unit digit and then we fill up the remaining places.
= 5 × 4 × 3 × 2 × 1 = 120
Hence, required number is 120.
6) How many even numbers greater than 300 can be formed with the digits 1, 2, 3, 4, 5 no digit being repeated?
(a) 111
(b) 600
(c) 900
(d) None of these
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(a) 111
Explanation:
Case 1
= 3 × 3 × 1 = 9
When 2 is fixed at unit place, we have 3, 4 and 5 i.e., 3 digits for hundreds place and remaining 3 digits (out of 5) for the tens place.
= 2 × 3 × 1 = 6
When 4 is fixed at unit place, we have 3 and 5 i.e., only two digits for hundreds place and remaining 3 digits for tens place.
Case 2
= 48
Case 3
= 48
5 digit numbers = 48
Thus there are total 15 + 48 + 48 = 111 even numbers greater than 300.
7) In how many ways the letters of the word RAINBOW be arranged?
(a) 5040
(b) 4050
(c) 3040
(d) 8040
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(a) 5040
Explanation:
There are 7 letters in the word RAINBOW and each letter is used only once. So all the 7 letters can be arranged in 7! Ways.
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
8) In the word RAINBOW, how many words begin with R?
(a) 720
(b) 360
(c) 1440
(d) None of these
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(a) 720
Explanation:
If we fix R as initial letter, then we have to arrange only 6 remaining letters.
Hence required number of permutations = 6!
= 6 × 5 × 4 × 3 × 2 × 1 = 720
9) In the word RAINBOW, how many words begin with R and ends with W?
(a) 120
(b) 240
(c) 180
(d) 360
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(a) 120
Explanation:
If we fix R and W as the first and last letters then we have to arrange only 5 remaining letters which can be arranged in 5! = 120 ways.
10) In the word RAINBOW, how many words are there in which I and O are never together?
(a) 1440
(b) 720
(c) 3600
(d) None of these
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(c) 3600
Explanation:
Number of permutations when I and O are together
= 2! × 6! = 1440
Total number of permutations = 7! = 5040
Therefore, number of permutations when I and O are not together
= 5040 – 1440 = 3600
