Permutation Test

Result

Permutation Test - 5

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The number of ways in which 10 different balls can be placed in three different boxes so that at least one box remains empty -

A
3(2¹⁰–2)

B
2¹⁰–1

C
3(2¹⁰–1)

D
None of these

Solution

No. of ways of choosing at least
one box of 3 boxes = ³C₁ = 3
10 different balls can be put in three boxes.
= ¹⁰C₁ + ¹⁰C₂ + ¹⁰C₃ + ¹⁰C₄ + … + ¹⁰C₁₀ = (2¹⁰ –1)
∴ using multiplication rule, we get
No. of required ways = 3 × (2¹⁰ – 1)
Question 2
1 / -0

The total number of ways in which a beggar can be given at least one rupee from four 25-paise coins, three 50-paise coins, and two one-rupee coins is

A
54

B
53

C
51

D
48

Solution

Four cases are there
(i) at least one one-rupee coin and any number of other coins is 2×4×5=40 ways.
(ii) at least two 50-paise coins and any number of 25-paise coins= 2x5=10 ways.
(iii) one 50-paise coin + at least two 25-paise coins = 1x3=3 ways
(iv) four 25-paise coins in one way only.

So, total number of method =40 + 10 + 3 + 1 =54.
Hence, the option A is correct.
Question 3
1 / -0

The sum of the factors of 8! which are odd and are of the form 3m + 2, where m is a natural number is

A
40

B
45

C
50

D
none of these

Solution

They are only 5 & 35, sum is 40
Hence, option A is correct.
Question 4
1 / -0

A box contains 6 balls which may be all of different colours or three each of two colours or two each of three different colours. The number of ways of selecting 3 balls from the box (if ball of same colour are identical), is-

A
60

B
31

C
30

D
none

Solution

When all 6 balls are different colour then ways for selecting 3 balls = ⁶C₃ = 20
when 3 balls one colour and 3 are other then ways selecting 3 balls = ²C₁¹C₁ + ²C₁ = 4
when three group of 2 balls each in same colour then ways for selecting three balls
= ³C₃ + ³C₁ . ²C₁ = 7
Total ways = 20 + 4 + 7 = 31 ways
Question 5
1 / -0

Number of ways in which 2 Indians, 3 Americans, 3 Italians and 4 Frenchmen can be seated on a circle, if the people of the same nationality sit together is-

A
2 . (4!)² ( 3!)²

B
2 . (3!)³ . 4!

C
2 . (3!) (4!)³

D
None of these

Solution

Let's consider them as 4 entities to be arranged in a circle, the number of ways for the same is 3!

each of those entities can be further arranged as follows:

2 Indians = 2!

3 Americans = 3!

3 Italians = 3!

4 Frenchmen = 4!

Hence total number of ways = 2