Test: Basic Concepts Of Permutations And Combinations- 1
40 Questions MCQ Test Quantitative Aptitude for CA CPT | Test: Basic Concepts Of Permutations And Combinations- 1
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Choose the most appropriate option (a) (b) (c) or (d)
4P3 is evaluated as
4P3
= 4!/(4!-3!)
= (4!)/1!
= 4*3*2*!
= 24
is equal to 
is a symbol equal to 0! = 1
In nPr , the restriction is
In nPr = n (n–1) (n–2) ………………(n–r–1), the number of factor is
If nP4 = 12 × nP2, the n is equal to
If . nP3 : nP2 = 3 : 1, then n is equal to
m+nP2 = 56, m–nP2 = 30 then
If 5Pr = 60, then the value of r is
If n1+n2P2 = 132, n1–n2P2 = 30 then,
The number of ways the letters of the word COMPUTER can be arranged is
Since all letters in the word "COMPUTER" are distinct then the arrangements is 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320
The number of arrangements of the letters in the word FAILURE, so that vowels are always coming together is
10 examination papers are arranged in such a way that the best and worst papers never come together. The number of arrangements is
No. of ways in which 10 paper can arrange is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10!- 9!.2! = 9!(10-2) = 8.9!.
n articles are arranged in such a way that 2 particular articles never come together. The number of such arrangements is
N. particles arrange in a way = n!
if two particles are arranged together then
there is only (n-1) particles
when two particles are together = (n-1)!
and two particles are arranged in= 2! ways
Thus, number of ways in which no particles
are not come together
=n! -2×(n-1)!
=( n-1)!×(n-2)
If 12 school teams are participating in a quiz contest, then the number of ways the first, second and third positions may be won is
The sum of all 4 digit number containing the digits 2, 4, 6, 8, without repetitions is
The number of 4 digit numbers greater than 5000 can be formed out of the digits 3,4,5,6 and 7(no. digit is repeated). The number of such is
4 digit numbers to be formed out of the figures 0, 1, 2, 3, 4 (no digit is repeated) then number of such numbers is
The number of ways the letters of the word “Triangle” to be arranged so that the word ’angle’ will be always present is
If the letters word ‘Daughter ’ are to be arranged so that vowels occupy the odd places, then number of different words are
There are only 3 vowels, a, u, and e, and there are 4 odd places. So
one of the odd spaces must contain a consonant.
We can choose 3 of the 4 odd places to put the vowels in C(4,3) or 4 ways.
For each of those vowel , we can rearrange them in P(3,3) or 3! or 6 ways.
And we can rearrange the 5 consonants in P(5,5) or 5! or 120 ways.
Hence the total number of different words formed are 6*4*120 = 2880
The number of ways in which 7 girls form a ring is
The number of ways in which 7 boys sit in a round table so that two particular boys may sit together is
No. of ways in which n people can sit in a round table = (n-1)!
So to make them sit together, consider them as a single unit in the sitting arrangement ie. n= 6.
Number of ways so that two particular boys may sit together in the round table:-
(6 -1)!*2
(The two particular boys can also be switched among themselves)
=> 120*2
=> 240
In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?
No. of ways in which 10 paper can arranged is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10! - 9! × 2!
= 9!(10 - 2)
= 8 × 9!
3 ladies and 3 gents can be seated at a round table so that any two and only two of the ladies sit together. The number of ways is
The number of ways in which the letters of the word DOGMATIC can be arranged is
The number of arrangements of 10 different things taken 4 at a time in which one particular thing always occurs is
Number of different things = 10
Numbers of things which are taken at a time = 4
Now, it is said that one particular thing always occur.
So The question reduces to selecting 9 things from the 3 things
This can be done easily by using the concept of combinations
Number of ways selecting 3 things out of 9 different things such that one particular thing always occur =
Now, the 4th thing is fixed so number of ways for selecting the 4th thing = 4! = 24
So, Required number of ways = 84 × 24 = 2016
The number of permutations of 10 different things taken 4 at a time in which one particular thing never occurs is
Mr. X and Mr. Y enter into a railway compartment having six vacant seats. The number of ways in which they can occupy the seats is
The number of numbers lying between 100 and 1000 can be formed with the digits 1, 2, 3, 4, 5, 6, 7 is
The number of numbers lying between 10 and 1000 can be formed with the digits 2,3,4,0,8,9 is
Two digits number (0 can't come on unit place) 5 × 5 = 25
Three digits number 5 × 5 × 4 = 100
Hence the total number of numbers between 10 and 1000 that can be formed using given digits is = 125
In a group of boys the number of arrangement of 4 boys is 12 times the number of arrangements of 2 boys. The number of boys in the group is
The total number of 9 digit numbers of different digits is
The number of ways in which 6 men can be arranged in a row so that the particular 3 men sit together, is
There are 5 speakers A, B, C, D and E. The number of ways in which A will speak always before B is
There are 10 trains plying between Calcutta and Delhi. The number of ways in which a person can go from Calcutta to Delhi and return by a different train is
The number of ways in which 8 sweets of different sizes can be distributed among 8 persons of different ages so that the largest sweat always goes to be younger assuming that each one of then gets a sweet is
The number of ways in which 8 sweets of different sizes can be distributed among 8 persons of different ages so that the largest sweat goes to the younger assuming that each one of them gets a sweet.
Since the largest goes to the youngest, that leaves 7 sweets to distribute to 7 people.
7P7 = 7! = 5040 ways
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