1.

In how many different number of ways 4 boys and 2 girls can sit on a bench?

**Answer: Option ''**

**Answer: Option 'C'
^{n}p_{n} = n!
^{6}p_{6}
= 6 × 5 × 4 × 3 × 2 × 1 = 720**

2.

In how many different number of ways 5 men and 2 women can sit on a shopa which can accommodate persons?

**Answer: Option 'D'**

**Answer: Option 'D' **

^{7}p_{3} = 7 × 6 × 5 = 210

3.

In how many different number of ways 4 boys and 3 girls can sit on a bench such that girls always sit together.

**Answer: Option 'A'**

**Answer: Option 'A'**

4.

In how many different ways can the letters of the word "CLAIM" be rearrangement?

**Answer: Option 'A'**

**Answer: Option 'A'
The total number of arrangements is
**

5.

If the letters of the word PLACE are arranged taken all at a time, find how many do not start with AE.

**Answer: Option 'C'**

**Answer: Option 'C'
Total no'of arrangements ^{5}P_{5} = 5! = 120
no'of arrangements start with AE = 1 × 6 = 6
no'of arrangements which do not start with AE = 120 - 6 = 114.**

6.

How many arrangements of the letters of the word BEGIN can be made, without changing the place of the vowels in the word?

**Answer: Option ''**

**Answer: Option 'B'
E,I fixed. Consonants can be arrangements in ^{3}P_{3} = 3! = 6 ways**

7.

If all the numbers 2, 3, 4, 5, 6, 7, 8 are arranged, find the number of arrangements in which 2, 3, 4, are together?

**Answer: Option 'A'**

**Answer: Option 'A'
If (2 3 4) is one.
we must arrange (2 3 4), 5, 6, 7, 8 in
**

8.

Find ^{10}P_{6}

**Answer: Option ''**

**Answer: Option 'B'
^{10}P_{6} = 10!/4! = 10 × 9 × 8 × 7 × 6 × 5
= 151200.**

9.

Find ^{9}P_{3}

**Answer: Option 'B'**

^{9}P_{3} = 9!/6! = 9 × 8 × 7

= 504.

10.

Find ^{7}P_{7}

**Answer: Option 'B'**

^{7}P_{7} = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

11.

Find ^{4}C_{2}?

**Answer: Option 'B'**

^{4}C_{2}

= 4!/(4-2)!2!

= 4!/(2! × 2!)

= (4 × 3)/2 = 6.

12.

Find ^{102}C_{99}?

**Answer: Option 'B'**

^{102}C_{99}

= 102!/(102-99)!99!

= 102!/(3! × 99!)

= (102 × 101 × 100)/(3 × 2)

34 × 101 × 50 = 171700.

13.

In how many different number of ways the letters of the word 'ENGINEERING' can be arranged.

**Answer: Option 'C'**

**11!/(3! × 3! × 2! × 2! × 1!)
= (11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)/(3 × 2 × 3 × 2 × 2 × 2 × 1)
= 11 × 10 × 8 × 7 × 5 × 3
= 277200**

14.

In how many different number of ways a combination of 3 persons can be selected from 4 men and 2 women.

**Answer: Option 'B'**

^{6}C_{3} × ^{5}C_{2}

^{6}C_{3}

= 6!/(3! . 3!)

= (6 × 5 × 4)/(3 × 2)

= 5 × 4 = 20.

^{5}C_{2}

= 5!/(3! . 2!)

= 5 × 2 = 10

= 20 × 10 = 200.

15.

In how many different number of ways a combination of 3 men and 2 women can be selected from 6 men and 5 women.

**Answer: Option 'B'**

^{6}C_{3}

= 6!/(3! . 3!)

= (6 × 5 × 4)/(3 × 2)

= 5 × 4 = 20.

16.

In how many different number of ways a Committee of 3 person of can be selected from 5 men and 3 women such that atleast 1 women is included in the committee.

**Answer: Option 'B'**

**1W 2M 2W 1M 3W
= ^{3}C_{1} × ^{5}C_{2} + ^{3}C_{2} × ^{5}C_{1} + ^{3}C_{3s}
= 3 × (5 × 4)/2 + 3 × 5 + 1
= 30 + 15 + 1 = 46
Total 5M 3W
^{8}C_{3} = 56
^{5}C_{3} = 10
Atleast one women = total - with out women
Atleast one women = 56 - 10 = 46**

17.

In how many different ways can the letters of the word 'SCHOOL' be arranged so that the vowels always come together?

**Answer: Option 'C'**

**'SCHOOL' contains 6 different letters.
vowels OO are always together
we have to arrange the letters (OO)SCHL
Now 5 letters can be arranged in 5! = 120 ways
The vowels (OO) can be arranged 2! = 2 ways.
= (120 x 2) = 240
again you have to divided 2 common vowels so answer is 120.**

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