26-08-2021 Answers

 1. How many 3-digit numbers can be formed from the digits 1, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Answer: 20

Solution: To be divisible by 5, 5 should be in the unit digit (as no zero is given).

Hence number of ways = 5*4*1 = 20

2. If the letters of the word 'MOTHER' are written in all possible order and these words are written out as in dictionary, find the rank of the word 'MOTHER'.

Answer: 309

Solution: The letters in alphabetical order are E H M O R T.

Words beginning with E and H = 120+120 = 240

Words beginning with ME and MH are = 24+24 = 48

Words beginning with MOE, MOH, MOR = 6+6+6 = 18

Words beginning with MOTE = 2

Hence rank of MOTHER = 240+48+18+2 + 1 = 309

3. In how many ways can 9 books be divided equally among 3 boys?

Answer: 12!/(24^3)

Solution: Number of ways of distributing 12 distinct items, into 3 groups of size 4 = 12!/((4!)^3)

= 12!/(24^3)

4. In a group of 6 men and 4 women, four people are to be selected. In how many different ways can they be selected such that at least one man should be there?

Answer: 209

Solution: Total num of ways to select 4 from 10 people = 10C4 = 210.

Num of ways to select all women = 4C4 =1.

Hence required num of ways = 210-1 = 209

5. In how many different ways can the letters of the word 'RETAIL' be arranged in such a way that the vowels occupy only the odd positions?

Answer: 36

Solution: 3 vowels are there out of 6 letters.

Num of odd positions = 3.

Hence ways to arrange vowels = 3C3*3! = 6

Ways to arrange consonants in remaining 3 even places= 3! = 6

Total ways = 6*6 = 36

6. In  how many ways can the letters of the word “SECTION” can be arranged so that no two vowels are together?

Answer: 1440

Solution: 3 vowels are present out of 7 letters.

_ C _ C _ C _ C _

The consonants can be arranged in these 4 slots in 4! ways.

The vowels can be arranged in any of the 5 remaining slots.

Hence total ways = 5C3 * 3!

Overall ways = 4! * 5C3 * 3!

= 4! * (5*4*3)/(3*2*1) * 3*2

= 1440

7. How many digits between 5000 and 7000 can be formed using the digits 3 to 8, when any digit can occur any number of times?

Answer: 432

Solution: From 3 to 8 we have 6 digits.

The thousandth digit can have either 5 or 6.

Hence number of ways = 2*6*6*6 = 432

8. How many digits between 6000 and 7000 can be formed using the digits 2 to 8, when any digit can occur any number of times?

Answer: 343

Solution: From 2 to 8 we have 7 digits.

The thousandth digit can have only 6.

Hence num of ways = 1*7*7*7 = 343

9. In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

Answer: 63

Solution: Num of ways = 7C5 * 3C2

= 7C2 * 3C1 = 63

10. There are 21 identical English books and 19 identical French Books. How can these books be arranged on a shelf where no 2 french books are placed together?

Answer: 1540

Solution: _ E _ E _ E _ ..... E _

Following the above arrangement, 43 slots are possible in which 22 are available for the French books to be placed.

Hence num of ways = 22C19 = 22C3 = 1540 ways

11. In how many ways 6 bikes can be parked in 9 vacant parking slots in a row?

Answer: 60480

Solution: The 6 parking slots can be chosen in 9C6 ways.

The 6 bikes can be arranged in 6! ways.

Total ways = 9C6 * 6! = 84 * 720 = 60480

12. If a,b,c are natural numbers, then how many solutions are there for a+b+c = 21 ?

Answer: 190

Solution: Here n=21 and r=3 (as there are 3 variables namely a,b,c).

Num of solutions possible = n-1 C r-1 = 20C2 = 190

13. There are 16 points in a plane of which 7 are in a straight line. Then how many straight lines can be formed?

Answer: 100

Solution: To make a straight like 2 points are needed.

Hence total num of straight lines = 16C2.

But as 7 points are already in a straight line 7C2 ways are lost and instead of them only one straight line is possible.

Hence effective straight lines = 16C2 - 7C2 + 1 = 120 - 21 + 1 = 100

14. In how many ways can 6 letters be placed in 4 different letter boxes?

Answer: 4096

Solution: As there are no restrictions like no box must be empty,

the number of ways = 4^6 = 4096

15. In  how many ways can the letters of the word “ANSWER” can be arranged so that the vowels are always together

Answer: 240

Solution: 2 vowels are present out of 6 letters.

If vowels are together effectively it is 5 letters with 2! ways to arrange the vowels among themselves.