1.
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
5P5 = 5! = 120 ways
2, 3, 4 can be arranged in 3P3 = 3! = 6
120 × 6 = 720.
2.
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.
3.
Find 102C99?
Answer: Option 'B'
102C99
= 102!/(102-99)!99!
= 102!/(3! × 99!)
= (102 × 101 × 100)/(3 × 2)
34 × 101 × 50 = 171700.
4.
Find 4C2?
Answer: Option 'B'
4C2
= 4!/(4-2)!2!
= 4!/(2! × 2!)
= (4 × 3)/2 = 6.
5.
In how many different number of ways 4 boys and 2 girls can sit on a bench?
Answer: Option ''
Answer: Option 'C'
npn = n!
6p6
= 6 × 5 × 4 × 3 × 2 × 1 = 720
6.
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
7.
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
= 3C1 × 5C2 + 3C2 × 5C1 + 3C3s
= 3 × (5 × 4)/2 + 3 × 5 + 1
= 30 + 15 + 1 = 46
Total 5M 3W
8C3 = 56
5C3 = 10
Atleast one women = total - with out women
Atleast one women = 56 - 10 = 46
8.
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'
9.
Find 7P7
Answer: Option 'B'
7P7 = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
10.
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'
7p3 = 7 × 6 × 5 = 210
11.
In how many different number of ways a combination of 3 persons can be selected from 4 men and 2 women.
Answer: Option 'B'
6C3 × 5C2
6C3
= 6!/(3! . 3!)
= (6 × 5 × 4)/(3 × 2)
= 5 × 4 = 20.
5C2
= 5!/(3! . 2!)
= 5 × 2 = 10
= 20 × 10 = 200.
12.
Find 9P3
Answer: Option 'B'
9P3 = 9!/6! = 9 × 8 × 7
= 504.
13.
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
5P5 = 5! = 120
14.
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'
6C3
= 6!/(3! . 3!)
= (6 × 5 × 4)/(3 × 2)
= 5 × 4 = 20.
15.
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 5P5 = 5! = 120
no'of arrangements start with AE = 1 × 6 = 6
no'of arrangements which do not start with AE = 120 - 6 = 114.