- 1
- 2
- 3
- …
- 7
- Next Page »

A Learning Portal from Recruitment India

From a pack of 52 cards, 3 cards are drawn together at random, What is the probability of both the cards are a king?

A.

1/5225

B.

1/5525

C.

5525

D.

1/525

Answer with explanation

Answer: Option BExplanation

Let S be the sample space

n(S) = 52C_{3} = [( 52 x 51 x 50) / (3 x 2 x 1)] = 13600 / 6 = 22100

Let E = event of getting 3 kings out of 4.

n(E) = 4C_{3} = [(4 x 3 x 2 x 1) / (3 x 2 x 1)] = 24/6 = 4

therefore, p(E) = [n(E) / n(S)]

==> p(E) = [ 4 / 22100]

==> p(E) = 1/5525

Workspace

16 persons are participated in a party. In how many differentways can they host the seat in a circular table, if the 2particular persons are to be seated on either side of the host?

A.

16! × 2

B.

14! × 2

C.

18! × 2

D.

14!

Answer with explanation

Answer: Option BExplanation

(16 – 2)! × 2 = 14! × 2

Workspace

How many 3-digit numbers can be formed from the digit 2,3,5,6,7 and 9, which are divisible by 5 and none of the digits is repeated?

A.

5

B.

10

C.

15

D.

20

Answer with explanation

Answer: Option DExplanation

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

Workspace

There is meeting of 20 delegates that is to be held in a hotel. In how many ways these delegates can be seated along a round table, if three particular delegates always seat together?

A.

17! 3!

B.

18! 3!

C.

D.

Answer with explanation

Answer: Option AExplanation

Total 20 persons, 3 always seat together, 17 + 1 =18 delegates can be seated in (18 -1)! Ways = 17!

And now that three can be arranged in 3! Ways.

So, 17! 3! is the correct answer.

Workspace

Groups each containing 3 boys are to be formed out of 5 boys. A, B, C, D and E such that no group can contain both C and D together. What is the maximum number of such different groups?

A.

5

B.

6

C.

7

D.

8

Answer with explanation

Answer: Option CExplanation

Maximum number of such different groups = ABC, ABD,ABE, BCE,BDE,CEA,DEA =7.

Alternate method:

Total number of way in which 3 boys can be selected out of 5 is 5C_{3}

Number of ways in which CD comes together = 3 (CDA,CDB,CDE)

Therefore, Required number of ways = 5C_{3} -3

= 10-3 =7.

Workspace

Find the number of ways of arranging the letters of the word “MATERIAL” such that all the vowels in the word are to come together?

A.

720

B.

1440

C.

1860

D.

2160

Answer with explanation

Answer: Option BExplanation

n the word, “MATERIAL” there are three vowels A, I, E.

If all the vowels are together, the arrangement is MTRL’AAEI’.

Consider AAEI as one unit. The arrangement is as follows.

M T R L A A E I

The above 5 items can be arranged in 5! ways and AAEI can be arranged among themselves in 4!/2! ways.

Number of required ways of arranging the above letters = 5! * 4!/2!

= (120 * 24)/2 = 1440 ways.

Workspace

In how many different ways can the letters of the word ‘OPTICAL’ be arranged so that the vowels always come together?

A.

120

B.

523

C.

560

D.

720

Answer with explanation

Answer: Option DExplanation

The word ‘OPTICAL’ contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Workspace

- 1
- 2
- 3
- …
- 7
- Next Page »

Correct Answer 👍

Wrong Answer 👎