IBPS PO Quantitative Aptitude: Permutation and Combination questions with solutions

Q1. Out of a total of 85 children playing badminton or table tennis, the total number of girls in the group is 70% of the total number of boys in the group. The number of boys playing only badminton is 50% of the number of girls, and the total number of boys playing badminton is 60% of the total number of boys. The number of children playing only table tennis is 40% of the total number of children, and a total of 12 children play both badminton and table tennis. What is the number of girls playing only badminton?

1. 16
2. 14
3. 17
4. None of these

Answer: None of these

The data can be modeled using two overlapping sets: badminton and table tennis. Using the total children, the boys-girls ratio, the number playing both games, and the given percentages, the count of girls playing only badminton does not match 16, 14, or 17. Hence the correct option is 'None of these'.

Q2. 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 included. In how many ways can it be done?

1. 564
2. 645
3. 735
4. 756

Answer: 756

At least 3 men means the committee can have 3 men 2 women, 4 men 1 woman, or 5 men 0 women. The number of ways is C(7,3)C(6,2) + C(7,4)C(6,1) + C(7,5)C(6,0) = 35×15 + 35×6 + 21×1 = 756.

Q3. In how many different ways can the letters of the word LEADING be arranged so that the vowels always come together?

1. 360
2. 480
3. 720
4. 5040

Answer: 720

LEADING has 7 distinct letters, with vowels E, A, I forming one block. So we arrange 5 units: (EAI), L, D, N, G in 5! ways, and the 3 vowels inside the block in 3! ways. Total = 5! × 3! = 720.

Q4. In how many different ways can the letters of the word CORPORATION be arranged so that the vowels always come together?

1. 810
2. 1440
3. 2880
4. 50400

Answer: 50400

CORPORATION has 11 letters: vowels O, O, A, I, O and consonants C, R, R, P, T, N. Treat the 5 vowels as one block, so we arrange 7 units: the vowel block and 6 consonants, with R repeated twice. Number of ways = 7!/2! × 5!/3! = 2520 × 20 = 50400.

Q5. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

1. 210
2. 1050
3. 25200
4. 21400

Answer: 25200

Choose 3 consonants from 7 and 2 vowels from 4, then arrange the 5 chosen letters. So the number of words = C(7,3) × C(4,2) × 5! = 35 × 6 × 120 = 25200.

Q6. In how many ways can the letters of the word "LEADER" be arranged?

1. 72
2. 144
3. 360
4. 720

Answer: 360

The word LEADER has 6 letters, with the letter E repeated twice. So the number of distinct arrangements is \(\frac{6!}{2!}\). This gives \(\frac{720}{2}=360\).

Q7. The information is about the total people who like three different games: volleyball, chess, and cricket. The people who like only volleyball are $(x+10)$, and the people who like only chess are 15 less than those who like only volleyball. The people who like only cricket are 28. The average number of people who like only one game is 21. The number of people who like all three games together is 50. The ratio of people who like volleyball and chess together to those who like chess and cricket together is 1:2. The total number of people who like chess is 96. Find the ratio of people who like only cricket and chess together to people who like only volleyball.

1. 24:25
2. 25:24
3. 15:16
4. 10:9

Answer: 24:25

The average of the three "only one game" groups is 21, so their total is 63. With only cricket = 28 and only chess being 15 less than only volleyball, the only volleyball count comes out to 25 and only chess to 10. Using the total chess count and the given overlap ratio, the required ratio simplifies to 24:25.

Q8. There are 350 students in a school who like Orange, Grapes, and Apple, and students can like more than one fruit as well. 6% of the total students like only Orange, 18% like only Apple, and 12% like only Grapes. $X\%$ of the total students like Orange and Grapes but not Apple, $Y\%$ like Apple and Orange but not Grapes, and $Z\%$ like Apple and Grapes but not Orange. The values of $X$, $Y$, and $Z$ are multiples of 10, and no two values are the same. How many students like all three fruits?

1. 28
2. 21
3. 14
4. 7

Answer: 14

6% of 350 = 21, 18% of 350 = 63, and 12% of 350 = 42. Let the pairwise-only groups be 10%, 20%, and 30% in some order, i.e. 35, 70, and 105 students; their sum is 210. Adding the single-fruit counts gives 21 + 63 + 42 = 126, so the remaining students are 350 - 126 - 210 = 14, who like all three fruits.

Q9. There are 14,000 students in a school. Out of these, 5,000 students like Maths, 4,500 like Science, and 4,000 like Sanskrit. There are 1,500 students who like all three subjects. There are 2,500 students who like Maths and Science, 2,700 who like Science and Sanskrit, and 2,300 who like Maths and Sanskrit. Some students do not like any subject. What is the ratio between the number of students who like at most two subjects and the number of students who like at most one subject?

1. 25: 19
2. 12:7
3. 23: 17
4. 2:1

Answer: 25: 19

Using inclusion-exclusion, we can determine the counts in each Venn region. The ratio of students who like at most two subjects to those who like at most one subject matches the option 25:19. This is a standard three-set Venn diagram problem.

Q10. The scatter chart given below shows the number of boys on the Y-axis and the number of girls on the X-axis in three different classes P, Q and R. Note: If there were 10 more girls in class R, then the probability of selecting a girl from R would be 40%. Q51. The number of girls in class P is 40% of the number of girls in R. If a teacher wants to make two groups, A and B, from classes R and P respectively such that in group A two girls and one boy are selected and in group B two girls and one boy are selected, find the difference between the number of ways he can make such groups from each class.

1. 23460
2. 22350
3. 21350
4. 22560

Answer: 23460

The probability condition gives the number of girls in class R, and then the number of girls in class P is 40% of that. After finding boys in each class from the scatter chart, the required number of ways is computed using combinations: \(\binom{g}{2}\binom{b}{1}\) for each class. The difference between the two counts is 23460.