6
Q:

# A Committee of 5 persons is to be formed from a group of 6 gentlemen and 4 ladies. In how many ways can this be done if the committee is to be included atleast one lady?

 A) 123 B) 113 C) 246 D) 945

Explanation:

A Committee of 5 persons is to be formed from 6 gentlemen and 4 ladies by taking.

(i) 1 lady out of 4 and 4 gentlemen out of 6

(ii) 2 ladies out of 4 and 3 gentlemen out of 6

(iii) 3 ladies out of 4 and 2 gentlemen out of 6

(iv) 4 ladies out of 4 and 1 gentlemen out of 6

In case I the number of ways = $C14×C46$ = 4 x 15 = 60

In case II the number of ways = $C24×C36$ = 6 x 20 = 120

In case III the number of ways = $C34×C26$ = 4 x 15 = 60

In case IV the number of ways = $C44×C16$ = 1 x 6 = 6

Hence, the required number of ways = 60 + 120 + 60 + 6 = 246

Q:

In how many ways the letters of the word 'CIRCUMSTANCES' can be arranged such that all vowels came at odd places and N always comes at end?

 A) 1,51,200 ways. B) 5,04,020 ways C) 72,000 ways D) None of the above

Explanation:

In circumcstances word there are 3C's, 2S's, I, U,R, T, A, N, E

Total = 13 letters

But last letter must be N

Hence, available places = 12

In that odd places = 1, 3, 5, 7, 9, 11

Owvels = 4

This can be done in 6P4 ways

Remaining 7 letters can be arranged in 7!/3! x 2! ways

Hence, total number of ways = 6P4 x 7!/3! x 2! = 360 x 5040/12 = 1,51,200 ways.

0 43
Q:

In how many different ways can the letters of the word 'RITUAL' be arranged?

 A) 720 B) 5040 C) 360 D) 180

Explanation:

The number of letters in the given word RITUAL = 6

Then,

Required number of different ways can the letters of the word 'RITUAL' be arranged = 6!

=> 6 x 5 x 4 x 3 x 2 x 1 = 720

2 157
Q:

How many four digits numbers greater than 6000 can be made using the digits 0, 4, 2, 6 together with repetition.

 A) 64 B) 63 C) 62 D) 60

Explanation:

Given digits are 0, 4, 2, 6

Required 4 digit number should be greater than 6000.

So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits.

This can be done by

1x4x4x4 = 64

Greater than 6000 means 6000 should not be there.

Hence, 64 - 1 = 63.

4 244
Q:

A card is drawn from a pack of 52 cards. What is the probability that either card is black or a king?

 A) 15/52 B) 17/26 C) 13/17 D) 15/26

Explanation:

Number of cards in a pack of cards = 52

Number of black cards = 26

Number of king cards = 4 (2 Red, 2 Black)

Required, the probability that if a card is drawn either card is black or a king =

4 352
Q:

A box contains 2 blue balls, 3 green balls and 4 yellow balls. In how many ways can 3 balls be drawn from the box, if at least one green ball is to be included in the draw?

 A) 48 B) 24 C) 64 D) 32

Explanation:

Total number of balls = 2 + 3 + 4 = 9

Total number of ways 3 balls can be drawn from 9 = 9C3

No green ball is drawn = 9 - 3 = 6 = 6C3

Required number of ways if atleast one green ball is to be included = Total number of ways - No green ball is drawn

= 9C3 - 6C3

= 9x8x7/3x2  -  6x5x4/3x2

= 84 - 20

= 64 ways.

4 450
Q:

How many 4-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

 A) 60 B) 48 C) 36 D) 20

Explanation:

Here given the required digit number is 4 digit.

It must be divisible by 5. Hence, the unit's digit in the required 4 digit number must be 0 or 5. But here only 5 is available.

x x x 5

The remaining places can be filled by remaining digits as 5 x 4 x 3 ways.

Hence, number 4-digit numbers can be formed are 5 x 4 x 3 = 20 x 3 = 60.

1 442
Q:

In how many ways the word 'SCOOTY' can be arranged such that 'S' and 'Y' are always at two ends?

 A) 720 B) 360 C) 120 D) 24

Explanation:

Given word is SCOOTY

ATQ,

Except S & Y number of letters are 4(C 2O's T)

Hence, required number of arrangements = 4!/2! x 2! = 4!

= 4 x 3 x 2

= 24 ways.

3 741
Q:

In how many ways word of 'GLACIOUS' can be arranged such that 'C' always comes at end?

 A) 3360 B) 5040 C) 720 D) 1080

Explanation:

Given word is GLACIOUS has 8 letters.

=> C is fixed in one of the 8 places

Then, the remaining 7 letters can be arranged in 7! ways = 5040.