## 10 Nov 2016

### Practice Sets on Quantitative Aptitude (Permutations and Combinations) | IBPS PO Mains Spl – Download in PDF

Practice Sets on Quantitative Aptitude (Permutations and Combinations) | IBPS PO Mains Spl – Download in PDF:
Dear Readers, Here we have given the important practice set questions on Quantitative Aptitude (Permutations and Combinations), aspirants those who are preparing for the examination can also download in pdf and make use of it.

Directions (1 – 2): There are 5 freshman, 8 sophomores and 7 juniors in a chess club. A group of 6 students will be chosen to compete in a competition.

1). How many combinations of students are possible if the group is to consist of exactly 3 freshmen?
a) 5000
b) 4500
c) 4550
d) 6250
e) 3500

2). How many combinations of students are possible if the group is to consist of all members of the same class?
a) 20
b) 25
c) 30
d) 35
e) 40

3). From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
a) 564
b) 645
c) 735
d) 756
e) None of these

4). Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
a) 24400
b) 21300
c) 21000
d) 25200
e) 23600

5). From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there in the committee. In how many ways can it be done?
a) 624
b) 702
c) 756
d) 812
e) 796

6). In how many different ways can the letters of the word 'MATHEMATICS' be arranged such that the vowels must always come together?
a) 9800
b) 100020
c) 120960
d) 140020
e) 152600

7). A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw?
a) 64
b) 128
c) 32
d) 256
e) 512

8). There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period?
a) 1800
b) 3200
c) 3600
d) Can’t be determined
e) None of these

9). A question paper has two parts P and Q, each containing 10 questions. If a student needs to choose 8 from part P and 4 from part Q, in how many ways can he do that?
a) 6020
b) 1200
c) 9450
d) Can’t be determined
e) None of these

10). In how many ways can 5 man draw water from 5 taps if no tap can be used more than once?
a) 60
b) 120
c) 240
d) 480
e) 720

1)b   2)d   3)d   4)d   5)c   6)c   7)a   8)a   9)c   10)b

Solutions:
1. B) Here we need the number of possible combinations of 3 out of 5 freshmen, 5C3 , and the number of possible combinations of 3 out of the 15 sophomores and juniors, 15C3.
Note that we want 3 freshmen and 3 students from the other classes. Therefore, we multiply the number of possible groups of 3 of the 5 freshmen times the number of possible groups of 3 of the 15 students from the other classes.
5C3 × 15C3 = 4550

2. D) This part of the problem seems similar to the first three parts but it is very different.
Here we want 6 freshmen or 6 sophomores or 6 juniors.
The group cannot be all freshmen since there are only 5 freshmen.
Therefore, the group can be a group of 6 sophomores or 6 juniors.
Number of groups of sophomores =8C6=28
Number of groups of juniors=7C6=7
Since we want the number of groups of 6 sophomores or 6 juniors, we want the sum of each of these possibilities:
=28+7=35

3. D)

4. D) Number of ways of selecting 3 consonants from 7 = 7C3
Number of ways of selecting 2 vowels from 4 = 4C2
Number of ways of selecting 3 consonants from 7 and 2 vowels from 4 = 7C3 × 4C2 = 210
It means we can have 210 groups where each group contains total 5 letters (3 consonants and 2 vowels).
Number of ways of arranging 5 letters among themselves = 5! = 5×4×3×2×1 = 120
Hence, required number of ways =210 × 120 = 25200

5. C)

6. C) The word 'MATHEMATICS' has 11 letters. It has the vowels 'A','E','A','I' in it and these 4 vowels must always come together. Hence these 4 vowels can be grouped and considered as a single letter. That is, MTHMTCS (AEAI).
Hence we can assume total letters as 8. But in these 8 letters, 'M' occurs 2 times, 'T' occurs 2 times but rest of the letters are different.
Hence, number of ways to arrange these letters = 8!/(2! * 2!) = 10080
In the 4 vowels (AEAI), 'A' occurs 2 times and rest of the vowels are different.
Number of ways to arrange these vowels among themselves = 4!/2! = 12
Hence, required number of ways = 10080 * 12 = 120960

7. A) From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that at least one black ball should be there.
Hence we have 3 choices as given below
We can select 3 black balls...(option 1)
We can select 2 black balls and 1 non-black ball ...(option 2)
We can select 1 black ball and 2 non-black balls ...(option 3)
Number of ways to select 3 black balls = 3C3
Number of ways to select 2 black balls and 1 non-black ball = 3C2 × 6C1
Number of ways to select 1 black ball and 2 non-black balls = 3C1 × 6C2
Total number of ways = 3C3 + 3C2 × 6C1 + 3C1 × 6C2 = 3C3 + 3C1 × 6C1 + 3C1 × 6C2 = 64

8. A)
5 subjects can be selected in 5C5 ways.
1 subject can be selected in 5C1 ways.
These 6 subjects can be arranged themselves in 6! ways.
Since two subjects are same, we need to divide by 2!
Therefore, total number of arrangements = 5C5 * 5C1 * 6!/2! = 1800

9. C) Number of ways to choose 8 questions from part P = 10C8
Number of ways to choose 4 questions from part Q = 10C4
Total number of ways= 10C8 × 10C4 10C2 × 10C4 = 45 * 210 = 9450

10. B) 1st man can draw water from any of the 5 taps.
2nd man can draw water from any of the remaining 4 taps.
3rd man can draw water from any of the remaining 3 taps.
4th man can draw water from any of the remaining 2 taps.
5th man can draw water from remaining 1 tap.
Hence total number of ways =5×4×3×2×1=120