28-08-2021 Answers

 1. How many terms are there in the expansionof (a+b+c)^10 ?

Answer: 66

Solution: This is like distributing 10 power among three groups.

Hence number of terms = n+r-1 C r-1 = 10+3-1 C 3-1 = 12C2 = 66

2. How many 5-digit numbers can be formed using the digits 1, 2, 3, 4 and 5 without repetition?

Answer: 120

Solution: Since all the five digits are distinct and all of these are used, the result should be 5! = 120.

3. If words are formed with letters in the word SHUVANK and arranged in alphabetical order in a dictionary, find the 55th word in the dictionary?

Answer: AHSNKUV

Solution: First arrange the letters in alphabetical order.

A H K N S U V

 

Now total words possible with each letter in the beginning is = (7-1)! = 720.

As 55 is less than 720, the starting letter is A.

 

Now 55 is more than 4! but less than 5!.

So AH will be the first two letters which will be followed by 120 combinations with KNSUV

 

Starting with AHK - we will have 24 words

Then starting with AHN - we will have 24 words

Now starting with AHSK - we will have 3! = 6 words. So till now 54 words have occurred.

Now 55th word will be AHSNKUV.

4. In how many ways 3 identical books may be distributed among 5 students such that no student gets more than 2 books?

Answer: 30

Solution: When 3 students get one book each and 2 dont get books, num of ways = 5C3 = 10 ways.

When one student gets one book and the second gets two books, number of ways = 5C1 * 4C1 = 20 ways. (After the first student is selected four remain)

So total ways = 10+20 = 30

5. In how many different ways can the letter of the word “BOARDING” be arranged in such a way that vowels always come together?

Answer: 4320

Solution: The word BOARDING contains 8 different letters

When the vowels OAI always come together, they can be supposed to form one letter.

Thus, we have to arrange the letters BRDNG (OAI).

Now, the 5 letters with (OAI) can be arranged in 6! ways = 720 ways

The vowels (OAI) canbe arranged in 3! ways = 6 ways

Required number of ways = 720 * 6 = 4320 ways

6. In how many ways can the letters of the word 'LEADER' be arranged?

Answer: 360

Solution: LEADER contains 6 letters 1L 2E 1A 1D and 1R.

So total number of arrangements that can be formed is = 6!/2! = 720/2 = 360.

7. If nC9 = nC11, what is the value of 22Cn ?

Answer: 231

Solution: n = 9+11 = 20

So 22C20 = 22C2 = 22*21/2 = 231

8. 4 couples have to sit around a circular table (with 8 seats) for dinner. In how many ways can they be seated so that men are not in adjacent positions?

Answer: 144

Solution: There are 4 men and 4 women.

Number of ways to place 4 men in circular fashion = (4-1)! = 3!.

Now 4 seats are available between these 4 men in which the 4 women can be seated in 4! ways.

Hence total num of ways = 3! * 4! = 6*24 = 144

9. A, B, C together own 4 carts and 6 bullocks. If cart needs only one bullock, find the number of different ways the can drive to a place, each in a different cart-bullock combination.

Answer: 2880

Solution: A can choose in 4C1 * 6C1 = 24 ways

B can choose in 3C1 * 5C1 = 15 ways

C can choose in 2C1 * 4C1 = 8 ways

Overall = 24*15*8 = 2880 ways

10. How many 4-digit numbers can be formed with the 10 digits 0,1,2,3,4,5,6,7,8,9 if the last digit must be 0 and repetition of digits is not allowed.

Answer: 504

Solution: The most significant digit cannot have zero (as it has no value). The last digit can have only zero.

Hence number of ways of forming the 4-digit number = 9 * 8 * 7 * 1 = 504

11. A man has 8 children. What is the number of ways in which he can take atleast 6 children to a restaurant?

Answer: 37

Solution: Atleast 6 means 6,7,8

Num of ways = 8C6 + 8C7 + 8C8

= 28+8+1 = 37

12. IT giant ABCD Consultancy lost its market value by 500 crores in stock exchange due to poor decision making process. There are 12000 employees working in the company. The board of directors decided to form a decision group out of 11 top managers to improve the performance of the company and regain the lost valuation. How many ways are there to form a group of 5 if the President and Vice President are not to serve on the same team?

Answer: 378

Solution: Scenario 1- Either VP or President is included.

4 places in the team needs to be filled from 9 people. 1 place needs to be filled from 2 people (VP or President). 

number of ways = 2* (9C4) = 2*126 = 252 

 

Scenario 2 - president and v president are not included: 

5 places in the team to be filled by 9 people. Number of ways = 9C5 = 126 

 

Total number of ways = 252+126 = 378

13. How many numbers from 1 to 2000 have none of their digits repeated?

Answer: 1242

Solution: 1 digit numbers = 9.

Two digit numbers = 9*9 = 81 (Zero cannot be in most significant digit)

Three digit numbers = 9*9*8 = 648

Four digit numbers = 1*9*8*7 = 504 (Only 1 can be in most significant digit)

Total = 9 + 81 + 648 + 504 = 1242

14. How many numbers formed out of the digits 4,5,6,8 lie between 4000 and 6000?

Answer: 12

Solution: The thousandth digit can be only 4 or 5 for the number to lie between 4000 and 6000.

Hence count = 2*3*2*1 = 12

15. Out of 9 consonants and 3 vowels, how many words (with or without meaning) of 3 consonants and 2 vowels can be formed?

Answer: 30240

Solution: Number of ways of selecting (3 consonants out of 9) and (2 vowels out of 3)

= 9C3 * 3C2 = 84*3 = 252 ways.

Again 5 letters can be arranged themselves in 5! = 120 ways.

So total number of ways = 252*120 = 30240

16. Amit has 6 toys. He wants to arrange them into 2 pairs, each consisting of three toys. In how many different ways can he arrange the toys?

Answer: 10

Solution: Num of ways of picking up first pair = 6C3.

Num of ways to pick second pair 3C3

So total number of ways = 6C3 * 3C3 = 20

But as the order of pair 1,2 does not matter, we divide it by 2!.

Hence required number of ways = (20)/(2*1) = 10

17. A group of 6 is chosen from 9 boys and 5 girls so as to contain at least 2 boys and 3 girls. How many different groups could be formed if two of the boys refuse to serve together?

Answer: 945

Solution: As the group of six has to contain atleast 2 boys and 3 girls, the possible combinations are

i) 2 boys 4 girls and ii) 3 boys 3 girls

Num of ways to pick 2 boys 4 girls = (9C2 - 1) x 5C4 = 35 x 5 = 175

Note: Subtract 1 since there is one combo of boys that are not allowed.

Num of ways to pick 3 boys 3 girls = (9C3 - 7) x 5C3 = (84-7) x 10 = 770

Note: Subtract 7 since there are 7 groups of three boys that can include two who refuse to work together. (9-2=7)

Adding these together we get 175+770 = 945.

18. From 5 different green balls, four different blue balls and three different red balls, how many combinations of balls can be chosen taking at least one green and one blue ball?

Answer: 3720

Solution: Here the number of balls being picked at a time is not mentioned. So the number of balls being picked at a given combination can be from 2 to 12 balls (minimum is 2 as atleast one green and one blue is desired)

Combination of atleast 1 green ball = 5C1+5C2+5C3+5C4+5C5 = 31 ways

Combination of atleast 1 blue ball = 4C1+4C2+4C3+4C4 = 15 ways

Combination of red balls = 3C0 + 3C1 + 3C2 + 3C3 = 8 ways

Total combination = 31*15*8= 3720 ways.

19. How many positive numbers can be formed with 2,4,5,9 without repetition?

Answer: 64

Solution: 1 digit numbers = 4

2 digit numbers = 4*3 = 12

3 digit numbers = 4*3*2 = 24

4 digit numbers = 4*3*2*1 = 24

Total = 64

20. A grandfather has 5 sons and daughters and 8 grandchildren. They have to be arranged in a row such that the first 4 seats and last four seats are to be taken by grandchildren and the grandfather would not sit adjacent to any of the grandchildren. How many possible ways are there to arrange?

Answer: 19353600

Solution: Overall 14 people (including grandfather) are to be seated. The seating arrangement should be like 

GC GC GC GC S/D _ _ _ _ S/D GC GC GC GC.

The eight grandchildren can occupy the seats denoted by GC in 8! ways. The grandfather can be seated in the blanks in 4 ways. 

The sons/daughters can occupy the two seats denoted by S/D and the remaining 3 seats denoted by blank(after grandfather occupies one seat among them) in 5! ways.

Hence required number of possible ways = 4 * 8! * 5! = 19353600