## Quant Quiz On Permutation & Combination Day 24 Bag

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1. How many 3 digit number can be formed with the digits 5, 6, 2, 3, 7 and 9 which are divisible by 5 and none of its digit is repeated?
a) 12
b) 16
c) 20
d) 24
e) None of these

Explanation :
_ _ 5
first two places can be filled in 5 and 4 ways respectively so, total number of 3 digit number = 5*4*1 = 20
2. In how many different ways can the letter of the word ELEPHANT be arranged so that vowels always occur together?
a) 2060
b) 2160
c) 2260
d) 2360
e) None of these

Explanation :
Vowels = E, E and A. They can be arranged in 3!/2! Ways
so total ways = 6!*(3!/2!) = 2160
3. There are 4 bananas, 7 apples and 6 mangoes in a fruit basket. In how many ways can a person make a selection of fruits from the basket.
a) 269
b) 280
c) 279
d) 256
e) None of these

Explanation :
Zero or more bananas can be selected in 4 + 1 = 5 ways (0 orange, 1 orange, 2 orange, 3 orange and 4 orange)
similarly apples can be selected in 7 +1 = 8 ways
and mangoes in 6 +1 = 7 ways
so total number of ways = 5*8*7 = 280
but we included a case of 0 orange, 0 apple and 0 mangoes, so we have to subtract this, so 280 – 1 = 279 ways
4. There are 15 points in a plane out of which 6 are collinear. Find the number of lines that can be formed from 15 points.
a) 105
b) 90
c) 91
d) 95
e) None of these

Explanation :
From 15 points number of lines formed = 15c2
6 points are collinear, number of lines formed by these = 6c2
So total lines = 15c2 – 6c2 + 1 = 91
5. In how many ways 4 Indians, 5 Africans and 7 Japanese be seated in a row so that all person of same nationality sits together
a) 4! 5! 7! 3!
b) 4! 5! 7! 5!
c) 4! 6! 7! 3!
d) can’t be determined
e) None of these

Answer – a) 4! 5! 7! 3!
Explanation :
4 Indians can be seated together in 4! Ways, similarly for Africans and Japanese in 5! and 7! respectively. So total ways = 4! 5! 7! 3!
6. In how many ways 5 Americans and 5 Indians be seated along a circular table, so that they are seated in alternative positions
a) 5! 5!
b) 6! 4!
c) 4! 5!
d) 4! 4!
e) None of these

Explanation :
First Indians can be seated along the circular table in 4! Ways and now Americans can be seated in 5! Ways. So 4! 5! Ways
7. 4 matches are to be played in a chess tournament. In how many ways can result be decided?
a) 27
b) 9
c) 81
d) 243
e) None of these

Explanation :
Every chess match can have three result i.e. win, loss and draw
so now of ways = 3*3*3*3 = 81 ways

Q(8 –9) There are 6 players in a cricket which is to be sent to Australian tour. The total number of members is 12.

1. If 2 particular member is always included
a) 210
b) 270
c) 310
d) 420
e) None of these

Explanation :
only 4 players to select, so it can be done in 10c4 = 210
2. If 3 particular player is always excluded
a) 76
b) 82
c) 84
d) 88
e) None of these

Explanation :
6 players to be selected from remaining 9 players in 9c6 = 84 ways
3. In a group of 6 boys and 5 girls, 5 students have to be selected. In how many ways it can be done so that at least 2 boys are included
a) 1524
b) 1526
c) 1540
d) 1560
e) None of these

Explanation :
6c2*5c3 + 6c3*5c2 + 6c4*5c1 + 6c5

• How many words of 4 letters with or without meaning be made from the letters of the word ‘NUMBER’, when repetition of letters is not allowed?
A) 480
B) 360
C) 240
D) 360
E) 24

D) 360
Explanation:

NUMBER is 6 letters.
We have 4 places where letters are to be placed.
For first letter there are 6 choices, since repetition is not allowed, for second, third and fourth letter also we have 5, 4, and 3 choices resp., so total of 6*5*4*3 ways = 360 ways.
• In how many ways the letters of the word ‘ALLIGATION’ be arranged taking all the letters?
A) 120280
B) 453600
C) 360340
D) 3628800
E) None of these

B) 453600
Explanation:

ALLIGATION contains 10 letters, so total 10! ways. There are 2 As, 2 Ls, 2 Is
So 10!/(2!*2!*2!)
• In how many ways all the letters of the word ‘MINIMUM’ be arranged such that all vowels are together?
A) 60
B) 30
C) 90
D) 70
E) 120

A) 60
Explanation:

Take vowels in a box together as one – IIU, M, N, M, M
So there are 5 that to be placed for this 5!, now 3 Ms, so 5!/3!, so arrangement of vowels inside box gives 3!/2!
So total = 5!/3! * 3!/2!
• In how many ways a group of 4 men and 3 women be made out of a total of 8 men and 5 women?
A) 720
B) 140
C) 120
D) 360
E) 210

B) 140
Explanation:

Total ways = 8C4*5C3
• How many 3 digit numbers are divisible by 4?
A) 256
B) 225
C) 198
D) 252
E) 120

B) 225
Explanation:

A number is divisible by 4 when its last two digits are divisible by 4
For this the numbers should have their last two digits as 00, 04, 08, 12, 16, … 96
By the formula, an = a + (n-1)d
96 = 0 + (n-1)*4
n = 25
so there are 25 choices for last 2 digits and 9 choices (1-9) for the 1st digit
so total 9*25
• How many 3 digits numbers have exactly one digit 2 in the number?
A) 225
B) 240
C) 120
D) 160
E) 185

A) 225
Explanation:

0 cannot be placed at first digit to make it a 3 digit number.
3 cases:
Case 1: 2 is placed at first place
1 choice for the first place, 9 choices each for the 2nd and 3rd digit (0-9 except 2)
So numbers = 1*9*9 = 81
Case 2: 2 is placed at second place
8 choices for the first place (1-9 except 2), 1 choice for the 2nd digit and 9 choices for the 3rd digit (0-9 except 2)
So numbers = 8*1*9 = 72
Case 3: 2 is placed at third place
8 choices for the first place (1-9 except 2), 9 choices for the 2nd digit (0-9 except 2) and 1 choice for the 3rd digit
So numbers = 8*9*1 = 72
So total numbers = 81+72+72 = 225
• There are 8 men and 7 women. In how many ways a group of 5 people can be made such that the particular woman is always to be included?
A) 860
B) 1262
C) 1001
D) 1768
E) 984

C) 1001
Explanation:

Total 15 people, and a particular woman is to be taken to form a group of 5, so choice is to be done from 14 people of 4 people
Ways are 14C4.
• There are 6 men and 7 women. In how many ways a committee of 4 members can be made such that a particular man is always to be excluded?
A) 280
B) 420
C) 220
D) 495
E) 460

D) 495
Explanation:

There are total 13 people, a particular man is to be excluded, so now 12 people are left to chosen from and 4 members to be chosen. So ways are 12C4.
• How many 4 digit words can be made from the digits 7, 8, 5, 0, and 4 without repetition?
A) 70
B) 96
C) 84
D) 48
E) 102

B) 96
Explanation:

0 cannot be on first place for it to be a 4 digit number,
So for 1st digit 4choices, for second also 4 (because 0 can be placed here), then 3 for third place, 2 for fourth place
Total numbers = 4*4*3*2
• In how many ways 8 students can be given 3 prizes such that no student receives more than 1 prize?
A) 348
B) 284
C) 224
D) 336
E) None of these

D) 336
Explanation:

For 1st prize there are 8 choices, for 2nd prize, 7 choices, and for 3rd prize – 6 choices left
So total ways = 8*7*6

• In how many ways can 3 prizes be given away to 12 students when each student is eligible for all the prizes ?
A.1234
B.1728
C.5314
D.1331
E.None of these

Explanation :
12^3 = 1728
• Total no of ways in which 30 sweets can be distributed among 6 persons ?
A.35 C 5
B.36 C 5
C.36 C 6
D.35!/5!
E.None of these

Explanation :
30+6-1 6-1 = 35 5
• A bag contains 4 red balls and 5 black balls. In how many ways can i make a selection so as to take atleast 1 red ball and 1 black ball ?
A.564
B.345
C.465
D.240
E.None of these

Explanation :
4-1 = 16 -1 = 15
5-1 = 32 -1 = 31
15*31 = 465
• In how many ways can 7 beads be strung into necklace ?
A.2520
B.5040
C.720
D.360
E.None of these

Explanation :
No of way in Necklace = (n-1)!/2 = 6!/2
= 720/2 = 360
• Find the no of 3 digit numbers such that  atleast one of the digit is 6 (with repetitions) ?
A.252
B.345
C.648
D.560
E.None of these

Explanation :
Total no of 3 digit number = 9*10*10 = 900
No of 3 digit number- none of the digit is 6 = 8*9*9 = 648
No of 3 digit number – atleast one digit is 6 = 900-648 = 252
• In how many ways  can 7 girls and 4 boys stand in a row so that no 2 boys are together ?
A.8467200
B.9062700
C.7407000
D.8407200
E.None of these

Explanation :
No of ways = 7!*8P4
7! = 5040
8P4 = 8*7*6*5 = 1680
No of ways = 5040*1680 = 8467200
• In how many ways the letters of the word PERMUTATION be arranged ?
A.10!/2!
B.10!
C.11!
D.11!/2!
E.None of these

Explanation :
No of ways = 11!/2!
• How many numbers can be formed with the digits 1, 7, 2, 5 without repetition ?
A.89
B.56
C.64
D.72
E.None of these