27-08-2021 Answers

1. There are 7 employees in a company and three are selected to do a task every day. This repeats as long as a distinct group can be formed, that is the combination of three employees in the group has never repeated (formed earlier). How many times can an employee be selected in a group?

Answer: 15

Solution: Let us consider any one employee X.

X can be paired with any two employees from the remaining 6.

Hence required num of times = 6C2 = 15

2. A football team consists of 16 players. It includes 2 strikers and 5 mid-fielders. In how many ways can a playiing eleven be selected if we have to select 1 forward keeper and at least 4 mid-fielders?

Answer: None of these

Solution: Ans: None of the above (As answer is 1092)

Num of ways of selecting a forward = 2C1 = 2.

Atleast 4 mid-fielders means it can be 4 or also 5.

When it is 4 mid-fielders,

Num of ways to pick the team = 2C1 * 5C4 * 9C6 = 2*5*84 = 840

When it is 5 mid-fielders,

Num of ways to pick the team = 2C1 * 5C5 * 9C5 = 2*1*126 = 252

Total ways = 840+252 = 1092

3. How many different numbers of 3 digits can be formed using only odd digits (1, 3, 5, 7, 9)?

Answer: 125

Solution: As nothing is mentioned we will assume repetition is allowed.

As it is a 3 digit number, unit digit can be filled in 5 ways.

Similarly tens and hundreds digit can be filled in 5 ways.

Hence required num of ways = 5*5*5 = 125

4. In how many ways can the letters A,B,C,D,E,F,G be arranged so that B and D are always adjacent to each other?

Answer: 1440

Solution: Treat B and D to be a single grouped letter. They can be arranged in 2! ways (BD and DB).

So overall possible number of ways = 6! * 2! = 1440

5. In how many ways can we arrange letters from A to Z, when A and I can have only seven letters between them?

Answer: 24! * 36

Solution: Out of the 26 slots, two slots are allocated for A and I and can be filled in 2,1 ways (The first slot can be A or I and the second slot the remaining).

The number of ways in which A and I can occur in different slots = 26-(7+1) = 18

Hence total num of ways = 24! * 2 * 18 = 24! * 36

6. Find the number of ways one can fill a 3*3 grid with 3 white marbles and 6 black marbles?

Answer: 84

Solution: The solution is just consider white or black marbles and calculate the possible ways they can be arranged in the grid.

Remaining positions are to be filled with the other color.

Hence number of ways considering white marbles = 9C3 = 84  (If you consider black too the answer will be 9C6 = 9C3 = 84)

7. In how many ways can LEADING be arranged in such a way that atleast two vowels always together?

Answer: 3600

Solution: Out of 7 we need 2 positions to intervene between vowels. In the remaining 5 , vowels can occupy in 5C3 ways and can be arranged among themselves in 3! ways. Remaining 4 places consonants in 4! ways. In this manner no vowel is together.

So ways atleast two together = 7! - (4! * (5C3*3!)) = 3600

8. In how many ways can 4 men and 3 ladies be arranged in a row so that no two men are together?

Answer: 144

Solution: M L M L M L M

The men be arranged in the slots marked M and the ladies can be arranged in the slots marked L.

Hence total num of ways = 4! * 3! = 144

9. There are atleast 5 identical red balls and 5 identical blue balls. 5 boxes are present and each box can hold only one ball. In how many possible patterns can the balls be distributed among the 5 boxes?

Answer: None of these

Solution: Each box can be filled in 2 ways (either red or blue).

Hence for 5 boxes, num of patterns = 2*2*2*2*2 = 32

10. A owes B  Rs. 250. He agrees to pay B over a number of consecutive days starting on a Monday, paying single note of Rs 50 or Rs 100 on each day. In how many different ways can A repay B. (Two ways are said to be different if at least one day, a note of a different denomination is given)

Answer: 8

Solution: Let us consider the possible count of Rs.50 notes being paid to pay Rs.250 and the pattern of payment across the days.

5 fifty rupee notes = 5C5 = 1

3 fifty rupee notes and 1 hundred rupee note = 4C1 = 4

1 fifty rupee note and 2 hundred rupee notes = 3C1 = 3

Hence total ways = 1+4+3 = 8

11. There are 60 people residing in a building out of which 60% are adults. A three member committee is to be formed choosing from the adults residing in the building. In how many ways can the committee be formed?

Answer: 7140

Solution: Num of adults = 60 * 0.6 = 36.

Num of ways of forming committee = 36C3 = 7140

12. In a box, there are 5 black pens, 3 white pens and 4 blue pens. In how many ways can 2 black pens, 2 white pens and 2 blue pens can be chosen?

Answer: 180

Solution: Num of ways = 5C2 * 3C2 * 4C2 = 180

13. 5 boys and 5 girls have to sit around a circular table (with 10 seats). In how many ways can they be seated so that boys are not in adjacent positions?

Answer: 2880

Solution: There are 5 boys and 5 girls.

Number of ways to place 5 boys in circular fashion = (5-1)! = 24

Now 5 seats are available between these 5 boys in which the 5 girls can be seated in 5! = 120 ways.

Hence total num of ways = 24 * 120 = 2880

14. There are 20 boys and 20 girls in a class. For staging a play we need one boy and one girl as lead pair. If a girl is too shy and refuses to act, how many lead pairs are possible?

Answer: None of these

Solution: As one girl refuses we have 19 girls to choose from.

Hence possible num of ways of picking boys= 20C1 = 20 and girls = 19C1 = 19

Hence required num of ways for lead pair = 20 * 19 = 380.

15. Three persons enter a bus with five vacant seats. In how many ways can they seat themselves?

Answer: 60

Solution: Num of ways of selecting 3 seats from five available vacant seats = 5C3 = 10

Num of ways of arranging three persons in the 3 seats selected = 3P3 = 6

Hence total number of ways to seat = 10*6 = 60

16. In a class there are 20 students. A group of 3 students is formed and will sing a song. This will continue till a distinct group of three students has never been formed earlier (That is the given 3 students have not formed a group earlier to sing a song). How many songs can be sung based on this condition?

Answer: 1140

Solution: This is nothing but the number of combinations of 3 students out of 20.

Hence num of songs sung = 20C3 = 1140

17. How many three digit odd numbers can you form such that if one of the digits is 2 then it is followed by 4?

Answer: 365

Solution: As number should be odd, the last digit can be 1,3,5,7,9.

Now there are two cases:

Case 1: The digit 2 is present. Now it should be followed by 4.

Hence number of possibilities = 1*1*5 = 5

Case 2: 2 is not present.

Now number of ways to fill unit digit = 5, tens digit = 9 (except 2 all can be used), hundreth digit = 8 (except 2 and 0).

Hence number of possibilities = 8*9*5 = 360

Total ways = 5+360 = 365

18. How many solutions are possible for the equation a + b + c = 50 so that a,b,c are whole numbers?

Answer: 1326

Solution: The formula to be used is (n+r-1)C(r-1).

Here n=50 and r=3.

Hence number of ways = 52C2 = 1326

19. There are 4 gates in a college. In how many ways can a person enter the college using a gate and leave by a different gate?

Answer: 12

Solution: Num of ways of choosing a gate for a given entry exit = 4C2 = 6

For a given combination of two gates, number of entry exit combination = 2P2 = 2

Hence total number of ways = 6*2 = 12

20. In how many ways can a pet shop owner arrange 5 dogs and 4 cats in a row so that no  two dogs are together?

Answer: None of these

Solution: D C D C D C D C D

The dogs can be arranged in the 5 slots marked D and the 4 cats can be arranged in the slots marked C.

Hence total num of ways = 5! * 4! = 120 * 24 = 2880