Permutation Test

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Permutation Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

Exponent of 12 in 50! is

A
24

B
22

C
19

D
21

Solution

12 = 2² × 3

Exponent of 12 in 50! = min{E₂(50!), E₃(50!)}

= E₃ (50!) = 22
Question 2
1 / -0

The number of zero's at the end 60! of is

A
17

B
16

C
14

D
12

Solution

Number of zero's at the end of 60! = exponent 10 in

60! = min{E₂ (60!), E₅ (60!)}

= E₅ (60!) = 14.
Question 3
1 / -0

Three dice are rolled. The number of possible outcomes in which at least one die shows 6 is :

A
91

B
201

C
125

D
48

Solution

Total number of outcomes = 6 × 6 × 6 = 216

Number of outcomes in which none of the dice shows 6 = 5 × 5 × 5 = 125

∴ Number of outcomes in which at least one die shows 6 = 216 - 125 = 91
Question 4
1 / -0

The maximum number of points of intersection of 8 circles, is

A
26

B
22

C
56

D
13

Solution

Two circles intersect in 2 points.

∴ Maximum number of points of intersection

= 2 × number of selection of two circles from 8 circles

= 2 × ⁸C₂ = 2 × 28

= 56.
Question 5
1 / -0

Let 1 ≤ m < n ≤ p. The number of subsets of the set A = {1, 2, 3,…p} having m,n as the least and the greatest elements respectively, is

A
None of these

B
2^{( n - m )}

C
2^{n - m - 1} - 1

D
2^{n - m - 1}

Solution

Total number of subject = the number of selections of at least two elements including m, n and natural numbers lying between m and n= total number of selections from n-m-1 different things 2^{n - m - 1}
Question 6
1 / -0

The number of odd proper divisors of 3^p. 6^m. 21ⁿ is

A
(p + m + n + 1) (n + 1) – 1

B
(p + 1) (m + 1) (n + 1) – 2

C
None of these

D
(p + 1) (m + 1) (n + 1) – 1

Solution

3^p. 6^m. 21ⁿ = 2^m. 3^{p + m + n}. 7ⁿ

∴ The required number of proper divisors

= Number of selection of any number of 3's and 7's

[∴ For odd divisors 2 must not be selected]

= (p + m + n + 1)(n + 1) - 1.
Question 7
1 / -0

The number of even proper divisors of 1008 is

A
23

B
24

C
22

D
None of these

Solution

1008 = 2⁴ × 3² × 7

∴ The required number of even proper divisors

= total number of selections of at least one 2 and any number of 3's or 7's

= 4 × (2 + 1) × (1 + 1) - 1 = 23.

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Re-Attempt Weekly Quiz Competition

Permutation Test - 10

Result

Permutation Test - 10

Score

-

Rank

-

SHARING IS CARING

Question 1

24

22

19

21

Solution

Question 2

17

16

14

12

Solution

Question 3

91

201

125

48

Solution

Question 4

26

22

56

13

Solution

Question 5

None of these

2( n - m )

2n - m - 1 - 1

2n - m - 1

Solution

Question 6

(p + m + n + 1) (n + 1) – 1

(p + 1) (m + 1) (n + 1) – 2

None of these

(p + 1) (m + 1) (n + 1) – 1

Solution

Question 7

23

24

22

None of these

Solution

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2^{( n - m )}

2^{n - m - 1} - 1

2^{n - m - 1}