Mathematics Multiple Choice Questions on “Permutations-1”.
1. Permutation is also known as selection.
a) True
b) False
View Answer
Answer: b
Clarification: Permutation is known as arrangement. Selection is another name for combinations.
It involves arrangement of letters, numbers, persons etc.
2. nPr = ________________
a) n!
b) (frac{n!}{r!})
c) (frac{n!}{(n-r)!})
d) (frac{n!}{(n-r)! r!})
View Answer
Answer: c
Clarification: Permutation is known as arrangement. nPr means arranging r objects out of n.
nPr = (frac{n!}{(n-r)!}).
3. 6! = _____________
a) 24
b) 120
c) 720
d) 8
View Answer
Answer: c
Clarification: We know, n! = n.(n-1).(n-2).(n-3)…..
6! = 6.5.4.3.2.1 = 720.
4. (frac{7!}{5!}) = ____________________
a) 7
b) 42
c) 230
d) 30
View Answer
Answer: b
Clarification: We know, n! = n.(n-1).(n-2).(n-3)…… = n(n-1)!
(frac{7!}{5!} = frac{7.6!}{5!} = frac{7.6.5!}{5!}) = 7.6 = 42.
5. (frac{100}{10!} = frac{1}{8!} + frac{x}{9!}). Find x.
a) 1
b) 2
c) 3
d) 4
View Answer
Answer: a
Clarification: (frac{100}{10!} = frac{1}{8!} + frac{x}{9!}).
We know, n! = n.(n-1). (n-2). (n-3) …………… = n(n-1)!
(frac{100}{10.9.8!} = frac{1}{8!} + frac{x}{9.8!})
=> (frac{100}{10.9} = frac{1}{1} + frac{x}{9})
=> (frac{10}{9} = 1 + frac{x}{9})
=> (frac{x}{9} = frac{10}{9} – 1 = frac{1}{9})
=> x=1.
6. nP0 = ________________
a) n!
b) 1
c) (frac{1}{(n)!})
d) (n-1)!
View Answer
Answer: b
Clarification: We know, nPr = (frac{n!}{(n-r)!}).
nP0 = (frac{n!}{(n-0)!} = frac{n!}{(n)!}) = 1.
7. nPn = ________________
a) n!
b) 1
c) (frac{1}{(n)!})
d) (n-1)!
View Answer
Answer: a
Clarification: We know, nPr = (frac{n!}{(n-r)!}).
nPn = (frac{n!}{(n-n)!} = frac{n!}{(0)!}) = n!
8. The number of permutations of n different objects taken r at a time, where repetition is allowed is _______________
a) n!
b) r!
c) nPr
d) nr
View Answer
1. Permutation is also known as selection.
a) True
b) False
View Answer
Answer: b
Clarification: Permutation is known as arrangement. Selection is another name for combinations.
It involves arrangement of letters, numbers, persons etc.
2. nPr = ________________
a) n!
b) (frac{n!}{r!})
c) (frac{n!}{(n-r)!})
d) (frac{n!}{(n-r)! r!})
View Answer
Answer: c
Clarification: Permutation is known as arrangement. nPr means arranging r objects out of n.
nPr = (frac{n!}{(n-r)!}).
3. 6! = _____________
a) 24
b) 120
c) 720
d) 8
View Answer
Answer: c
Clarification: We know, n! = n.(n-1).(n-2).(n-3)…..
6! = 6.5.4.3.2.1 = 720.
4. (frac{7!}{5!}) = ____________________
a) 7
b) 42
c) 230
d) 30
View Answer
Answer: b
Clarification: We know, n! = n.(n-1).(n-2).(n-3)…… = n(n-1)!
(frac{7!}{5!} = frac{7.6!}{5!} = frac{7.6.5!}{5!}) = 7.6 = 42.
5. (frac{100}{10!} = frac{1}{8!} + frac{x}{9!}). Find x.
a) 1
b) 2
c) 3
d) 4
View Answer
Answer: a
Clarification: (frac{100}{10!} = frac{1}{8!} + frac{x}{9!}).
We know, n! = n.(n-1). (n-2). (n-3) …………… = n(n-1)!
(frac{100}{10.9.8!} = frac{1}{8!} + frac{x}{9.8!})
=> (frac{100}{10.9} = frac{1}{1} + frac{x}{9})
=> (frac{10}{9} = 1 + frac{x}{9})
=> (frac{x}{9} = frac{10}{9} – 1 = frac{1}{9})
=> x=1.
6. nP0 = ________________
a) n!
b) 1
c) (frac{1}{(n)!})
d) (n-1)!
View Answer
Answer: b
Clarification: We know, nPr = (frac{n!}{(n-r)!}).
nP0 = (frac{n!}{(n-0)!} = frac{n!}{(n)!}) = 1.
7. nPn = ________________
a) n!
b) 1
c) (frac{1}{(n)!})
d) (n-1)!
View Answer
Answer: a
Clarification: We know, nPr = (frac{n!}{(n-r)!}).
nPn = (frac{n!}{(n-n)!} = frac{n!}{(0)!}) = n!
8. The number of permutations of n different objects taken r at a time, where repetition is allowed is _______________
a) n!
b) r!
c) nPr
d) nr
View Answer
Answer: d
9. Find the number of permutations of word DEPENDENT.
a) 132400
b) 1512500
c) 1663200
d) 1723400
View Answer
Answer: c
Clarification: There are total 9 letters out of which 1T, 2N, 2D, 3E, 1P.
Total number of permutations are (frac{9!}{3!2!2!} = frac{9.8.7.6.5.4.3.2.1}{6*2*2} = frac{362880}{24}) = 15120.
10. Find the number of 5 letter words which can be formed from word IMAGE without repetition using permutations.
a) 20
b) 60
c) 120
d) 240
View Answer
Answer: c
Clarification: IMAGE is a 5 letters word. We have to arrange all 5 letters of the word IMAGE without repetition. So, total permutations are nPr = 5P5 = 5! = 5.4.3.2.1 = 120.
11. Find the number of 5 letter words that can be formed from word IMAGE using permutations if repetition is allowed.
a) 25
b) 120
c) 125
d) 3125
View Answer
Answer: d
Clarification: IMAGE is a 5 letters word. We have to arrange all 5 letters of the word IMAGE with repetition allowed. So, total permutations are nr = 55 = 3125.
9. Find the number of permutations of word DEPENDENT.
a) 132400
b) 1512500
c) 1663200
d) 1723400
View Answer
Answer: c
Clarification: There are total 9 letters out of which 1T, 2N, 2D, 3E, 1P.
Total number of permutations are (frac{9!}{3!2!2!} = frac{9.8.7.6.5.4.3.2.1}{6*2*2} = frac{362880}{24}) = 15120.
10. Find the number of 5 letter words which can be formed from word IMAGE without repetition using permutations.
a) 20
b) 60
c) 120
d) 240
View Answer
Answer: c
Clarification: IMAGE is a 5 letters word. We have to arrange all 5 letters of the word IMAGE without repetition. So, total permutations are nPr = 5P5 = 5! = 5.4.3.2.1 = 120.
11. Find the number of 5 letter words that can be formed from word IMAGE using permutations if repetition is allowed.
a) 25
b) 120
c) 125
d) 3125
View Answer
Answer: d
Clarification: IMAGE is a 5 letters word. We have to arrange all 5 letters of the word IMAGE with repetition allowed. So, total permutations are nr = 55 = 3125.