Practice Permutations and combination - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) A boy has 3 library cards and 8 books of his interest in the library. Of these 8, he does not want to borrow chemistry part II unless Chemistry part I is also borrowed. In how many ways can he choose the three books to be borrowed ?
(a)
(b)
(c)
(d)
(e)
Two possibilities are there :
(i) Chemistry part I is available in 8 books with Chemistry part II.
or
(ii) Chemistry part II is not available in 8 books but Chemistry part I is available.
Total No. of ways = 1× 1 ×$^6C_1 + ^7C_3$ = 6 + ${7 × 6 × 5}/{3 × 2}$ = 6 + 35 = 41
Q-2) In how many different ways can be letters of the word 'CYCLE' be arranged?
(a)
(b)
(c)
(d)
(e)
CYCLE whereas C comes two times.
So, arrangements are = ${5!}/{2!}$ = ${5 × 4 × 3 × 2}/{2}$ = 60 ways
Q-3) The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by
(a)
(b)
(c)
(d)
(e)
6 men can be sit by 5! ways and on remaining 6 seats, 5 women can sit by $^6C_5$ ways.
∴ Required number of ways = 5! × $^6P_5$ = 6! × 5!
Q-4) In how many ways can six different rings be worn on four fingers of one hand ?
(a)
(b)
(c)
(d)
(e)
Required number of ways
= ways of selecting 4 objects out of 6 given objects
= $^6C_4$ = ${6 × 5}/2$ = 15
Q-5) The committee should consist of 2 Professors, 2 Teachers and 1 Reader?
(a)
(b)
(c)
(d)
(e)
Total number of ways to form committee
= $^5C_2 × ^6C_2 × ^3C_1$ = ${5 × 4}/{1 × 2}$ × ${6 × 5}/{1 × 2}$ × $3/1$= 10 × 15 × 3 = 450
Q-6) the committee should include all the 3 Readers?
(a)
(b)
(c)
(d)
(e)
Total number of ways to form committee
= $^5C_2 × ^6C_0 × ^3C_3 + ^5C_1 × ^6C_1 × ^3C_3 + ^5C_0 × ^6C_2 × ^3C_3$ = 10 + 30 + 15 = 55
Q-7) There are 4 candidates for the post of a lecturer in Mathematics and one is to be selected by votes of 5 men. The number of ways in which the votes can be given is
(a)
(b)
(c)
(d)
(e)
Let the four candidates gets the votes x, y, z and w such that
x + y + z + w = 51 ...(i)
Here x ≥ 0 , y ≥ 0, z ≥ 0, w ≥ 0
The number of solutions of the above equation in this case is same as the number of ways in which the votes can be given if at least no two candidates get equal number of votes.
(Note : The number of ways in which n identical things can be distributed into r different groups = $^{n + r – 1}{C_{r - 1}}$)
∴ Total number of solutions of eqn. (i)
= $^{5 + 4 – 1}C_{4 – 1}$ = $^8C_3$ = 56
But in 8 ways the two candidate gets equal votes which are shown below :
(2, 2, 1, 0), (2, 2, 0, 1), (0, 2, 2, 1), (1, 2, 2, 0), (0, 1, 2, 2), (1, 0, 2, 2), (2, 0, 1, 2), (2, 1, 0, 2)
Hence the required number of ways = 56 – 8 = 48
Q-8) In how many different ways can the letters of the word TRUST be arranged?
(a)
(b)
(c)
(d)
(e)
Required no. of ways = ${5!}/{2!}$ = 60 is
Total no. of letters in the word is 5; T is repeated twice.
Q-9) How many different ways can the letters in the word ATTEND be arranged?
(a)
(b)
(c)
(d)
(e)
There are 6 letter in the word 'ATTEND' whereas, T comes two times.
So, required number of ways = ${6!}/{2!}$ = ${720}/{2}$ = 360
Q-10) In how many different ways can the letters of the word 'ARMOUR' be arranged?
(a)
(b)
(c)
(d)
(e)
ARMOUR = 6 letter whereas R repeated twice
∴ ${6!}/{2!}$ = ${6 × 5 × 4 × 3 × 2 × 1}/{2 × 1}$ = 360
Q-11) In how many different ways can hte letters of the word 'PRETTY' be arranged?
(a)
(b)
(c)
(d)
(e)
Number of ways = ${6!}/{2!}$ (∵ T letter comes in two time)
= ${6 × 5 × 4 × 3 × 2 × 1}/{2 × 1}$ = 360
Q-12) Directions :
Answer these questions on the basis of the information given below : From a group of 6 men and 4 women a Committee of 4 persons is to be formed
In how many different ways can it be done, so that the committee has at least 2 men?
(a)
(b)
(c)
(d)
(e)
The committee of 4 persons is to be so formed that it has at least 2 men. The different ways that we can choose to form such a committee are:
(i) 2m. 2w in $^6C_2 × ^4C_2$ = ${6 × 5}/{2 × 1}$ × ${3 × 3}/{2 × 1}$ = 90
(ii) 3m. 1w in $^6C_3 × ^4C_1$ = ${6 × 5 × 4}/{3 × 2 × 1}$ × 4= 80
(iii) 4m in $^6C_4$ = ${6 × 5}/{2 × 1}$ = 15
∴ Total no. of different ways in which a committee of 4 persons can be formed so that it has at least 2 men.
= 90 + 18 + 15 = 185
Q-13) In how many ways can 5 boys be chosen from 6 boys and 4 girls so as to include exactly one girl?
(a)
(b)
(c)
(d)
(e)
One girl can be chosen in $^4C_1$ = 4 ways
and 4 boys can be chosen in $^6C_4$ = 15 ways
∴ Total number of ways = 4 × 15 = 60 ways
Q-14) A student is to answer 10 out of 13 questions in an examination such that he must choose at least four from the first five questions. The number of choices available to him is:
(a)
(b)
(c)
(d)
(e)
The student can answer by using the following combinations 4 from 5 and 6 from 8 or 5 from 5 and 5 from 8
i.e. $^5C_4 × ^8C_6 + ^5C_5 × ^8C_5$ = ${5!}/{4!1!}$ × ${8!}/{6!2!}$ + ${5!}/{5!0!}$ × ${8!}/{3!5!}$
= 140 + 56 = 196 Ways
Q-15) There are 100 students in a college class of which 36 are boys studying statistics and 13 girls not studying statistics. If there are 55 girls in all, then the probability that a boy picked up at random is not studying statistics, is
(a)
(b)
(c)
(d)
(e)
There are 55 girls and 45 boys in the college.
Out of 45 boys, 36 are studying Statistics and 9 are not studying statistics.
∴ The probability that a boy picked up at random is not
studying Statistics = $9/{45}$ = $1/5$
Q-16) A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
(a)
(b)
(c)
(d)
(e)
The student can choose 4 questions from first 5 questions or he can also choose 5 questions from the first five questions.
∴ No. of choices available to the student
= $^5C_4 × ^8C_6 + ^5C_5 × ^8C_5$ = 196.
Q-17) In how many ways can 7 persons be seated at a round table if 2 particular persons must not sit next to each other ?
(a)
(b)
(c)
(d)
(e)
Total no. of unrestricted arrangements = (7 – 1) ! = 6 !
When two particular person always sit together, the total no. of arrangements = 6! – 2 × 5!
Required no. of arrangements = 6! – 2 × 5!
= 5! (6 – 2) = 5 × 4 × 3 × 2 × 4 = 480.
Q-18) In how many different ways can 4 boys and 3 girls be arranged in a row such that all boys stand together and all the girls stand together?
(a)
(b)
(c)
(d)
(e)
Total number of ways to stand boys and girls together
= 4! × 3! × 2! = 4 × 3 × 2 × 3 × 2 × 2 = 288
Q-19) Directions :
Answer these questions on the basis of the information given below : From a group of 6 men and 4 women a Committee of 4 persons is to be formed
In how many different ways can it be done so that the committee has at least one woman?
(a)
(b)
(c)
(d)
(e)
The committee of 4 persons is to be so formed that it has at least 1 woman.
The different ways that we can choose to form such a committee are:
(i) 1w. 3 m in $^4C_1 × ^6C_3$ = 4 × ${6 × 5 × 4}/{3 × 2 × 1}$ = 80
(ii) 2w. 2 m in $^4C_2 × ^6C_2$ = ${4 × 3}/{2 × 1}$ × ${6 × 5}/{2 × 1}$ = 90
(iii) 3w. 1 m in $^4C_3 × ^6C_1$ = 4 × 6 = 24
(iv) 4w in $^4C_4$ = 1
∴ Total no. of different ways in which a committee of 4 persons can be formed so that it has at least one woman. = 80 + 90 + 24 + 1 = 195
Q-20) In how many different ways can the letters of the word 'PROBLEM' be arranged ?
(a)
(b)
(c)
(d)
(e)
The word PROBLEM consists of 7 distinct letters.
∴ Number of arrangements = 7!
= 70 × 6 × 5 × 4 × 3 × 2 × 1 = 5040