6 + Interview Questions in NUMERICAL ABILITY in PERMUTATION AND COMBINATION Page 1 InterviewSolution

1.	Quantity B: 30001). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B
Answer» Quantity A: Four gadgets can be chosen such that there is at LEAST one gadget of each type, if two gadgets of one of the type is chosen and one gadget of remaining two types is chosen. ∴ NUMBER of ways in which this can be DONE = 7C2 × 6C1 × 8C1 + 7C1 × 6C2 × 8C1+ 7C1 × 6C1 × 8C2 = 3024 Quantity B: 3000 ∴ Quantity A < Quantity B

1.

Quantity B: 30001). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B

Answer»

Quantity A:

Four gadgets can be chosen such that there is at LEAST one gadget of each type, if two gadgets of one of the type is chosen and one gadget of remaining two types is chosen.

∴ NUMBER of ways in which this can be DONE = 7C2 × 6C1 × 8C1 + 7C1 × 6C2 × 8C1+ 7C1 × 6C1 × 8C2 = 3024

Quantity B:

3000

∴ Quantity A < Quantity B

Discussion

2.	1). Quantity 1 > Quantity 2 2). Quantity 1 ≥ Quantity 2 3). Quantity 2 > Quantity 14). Quantity 2 ≥ Quantity 1
Answer» Quantity 1: If we fix 1 at the unit digit place than any other digit (7, 9) can be arranged in = 2! = 2 Which means there 2 numbers which have 1 at the unit place If we observed we will find that there are 2 numbers each with 7 and 9 at the unit place. SUM of all the numbers at the unit’s place is, ⇒ 1 × 2 + 7 × 2 + 9 × 2 ⇒ 2 × (1 + 7 + 9) ⇒ 2 × (17) = 34 Sum of all the tens place = 34 × 10 = 340 Sum of all the hundred place = 34 × 100 = 34000 ∴ Sum of all POSSIBLE numbers = 34 + 340 + 34000 = 34374 Quantity 2: If we fix 1 at the unit digit place than any other digit (8, 7) can be arranged in = 2! = 2 Which means there 2 numbers which have 1 at the unit place If we observed we will find that there 2 numbers each with 7 and 8 at the unit place Sum of all the numbers at the unit’s place, ⇒ 1 × 2 + 7 × 2 + 8 × 2 ⇒ 2 × (1 + 7 + 8) ⇒ 2 × (16) = 32 Sum of all the tens place = 32 × 10 = 320 Sum of all the hundred place = 32000 ∴ Sum of all possible number = 32 + 320 + 32000 = 32352 We can see that ∴ Quantity 1 > Quantity 2

2.

1). Quantity 1 > Quantity 2 2). Quantity 1 ≥ Quantity 2 3). Quantity 2 > Quantity 14). Quantity 2 ≥ Quantity 1

Answer»

Quantity 1:

If we fix 1 at the unit digit place than any other digit (7, 9) can be arranged in = 2! = 2

Which means there 2 numbers which have 1 at the unit place

If we observed we will find that there are 2 numbers each with 7 and 9 at the unit place.

SUM of all the numbers at the unit’s place is,

⇒ 1 × 2 + 7 × 2 + 9 × 2

⇒ 2 × (1 + 7 + 9)

⇒ 2 × (17) = 34

Sum of all the tens place = 34 × 10 = 340

Sum of all the hundred place = 34 × 100 = 34000

∴ Sum of all POSSIBLE numbers = 34 + 340 + 34000 = 34374

Quantity 2:

If we fix 1 at the unit digit place than any other digit (8, 7) can be arranged in = 2! = 2

Which means there 2 numbers which have 1 at the unit place

If we observed we will find that there 2 numbers each with 7 and 8 at the unit place

Sum of all the numbers at the unit’s place,

⇒ 1 × 2 + 7 × 2 + 8 × 2

⇒ 2 × (1 + 7 + 8)

⇒ 2 × (16) = 32

Sum of all the tens place = 32 × 10 = 320

Sum of all the hundred place = 32000

∴ Sum of all possible number = 32 + 320 + 32000 = 32352

We can see that ∴ Quantity 1 > Quantity 2

Discussion

3.	1). 14002). 10403). 21604). 720
Answer» In the word LUCKNOW, we TREAT the VOWELS UO as one letter. Thus we have, LCKNW (UO) total 6 LETTERS. There is no repetition of letters. ∴ Number of ways to ARRANGE these letters = 6! = 720 Now, the 2 vowels can be arranged is = 2! = 2 ways ∴ Total no. of arrangements = 720 × 2 = 1440 ways

3.

1). 14002). 10403). 21604). 720

Answer»

In the word LUCKNOW, we TREAT the VOWELS UO as one letter.

Thus we have, LCKNW (UO) total 6 LETTERS.

There is no repetition of letters.

∴ Number of ways to ARRANGE these letters = 6! = 720

Now, the 2 vowels can be arranged is = 2! = 2 ways

∴ Total no. of arrangements = 720 × 2 = 1440 ways

Discussion

4.	1). 3602). 3103). 2404). 550
Answer» We KNOW that, The first letter must ALWAYS be filled with the two vowels in the WORD The number of ways the 2 vowels can fill the first position = 2! = 2 The number of ways the REST of the letters fill the remaining 5 spaces can be filled 5! = 120 Total number of ways of ARRANGING the word = 120 × 2 = 240 ∴ the number of ways the word can be arranged = 240

4.

1). 3602). 3103). 2404). 550

Answer»

We KNOW that,

The first letter must ALWAYS be filled with the two vowels in the WORD

The number of ways the 2 vowels can fill the first position = 2! = 2

The number of ways the REST of the letters fill the remaining 5 spaces can be filled 5! = 120

Total number of ways of ARRANGING the word = 120 × 2 = 240

∴ the number of ways the word can be arranged = 240

Discussion

5.	In how many different ways can the letters of the word ‘SIMPLE’ be arranged?1). 5202). 1203). 50404). 270
Answer» The NUMBER of WAYS of arranging a word of letters n is n! SIMPLE has 6 letters. ∴ Number of ways of arranging the word SIMPLE = 6! = 720

5.

In how many different ways can the letters of the word ‘SIMPLE’ be arranged?1). 5202). 1203). 50404). 270

Answer»

The NUMBER of WAYS of arranging a word of letters n is n!

SIMPLE has 6 letters.

∴ Number of ways of arranging the word SIMPLE = 6! = 720

Discussion

6.	On a shelf there are 3 books of Chemistry, 2 books of Physics and 4 books of Mathematics. In how many different ways can be the books be arranged such that the books of Mathematics are together?1). 3628802). 86403). 345604). 17280
Answer» SINCE the BOOKS of Mathematics are to be kept TOGETHER, they are considered as 1 ∴ Total number of books = Group of Mathematics books + Physics + Chemistry ⇒ Total books = 1 + 2 + 3 = 6 Number of ways in which all books can be arranged = 6! Number of ways in which the Mathematics books can be arranged = 4! ∴ Total number of ways in which the books can arranged = 6! × 4! = 17280

6.

On a shelf there are 3 books of Chemistry, 2 books of Physics and 4 books of Mathematics. In how many different ways can be the books be arranged such that the books of Mathematics are together?1). 3628802). 86403). 345604). 17280

Answer»

SINCE the BOOKS of Mathematics are to be kept TOGETHER, they are considered as 1

∴ Total number of books = Group of Mathematics books + Physics + Chemistry

⇒ Total books = 1 + 2 + 3 = 6

Number of ways in which all books can be arranged = 6!

Number of ways in which the Mathematics books can be arranged = 4!

∴ Total number of ways in which the books can arranged = 6! × 4! = 17280

Discussion

Explore topic-wise InterviewSolutions in .

Quantity B: 30001). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B

1). Quantity 1 > Quantity 2 2). Quantity 1 ≥ Quantity 2 3). Quantity 2 > Quantity 14). Quantity 2 ≥ Quantity 1

1). 14002). 10403). 21604). 720

1). 3602). 3103). 2404). 550

In how many different ways can the letters of the word ‘SIMPLE’ be arranged?1). 5202). 1203). 50404). 270

On a shelf there are 3 books of Chemistry, 2 books of Physics and 4 books of Mathematics. In how many different ways can be the books be arranged such that the books of Mathematics are together?1). 3628802). 86403). 345604). 17280