Permutation MCQ Question with Answer

Correct Answer : C

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Explanation

Number of ways of selecting 3 consonants out of 7 = ⁷C₃
Number of ways of selecting 2 vowels out of 4 = ⁴C₂
Number of ways of selecting 3 consonants out of 7 and 2 vowels out of 4 = ⁷C₃ x ⁴C₂

$$\begin{gathered} =\left(\frac{7\times 6\times 5}{3\times 2\times 1}\right)\times \left(\frac{4\times 3}{2\times 1}\right) \\ =210 \end{gathered}$$

It means that we can have 210 groups where each group contains total 5 letters(3 consonants
and 2 vowels).
Number of ways of arranging 5 letters among themselves = 5!
= 5 x 4 x 3 x 2 x 1 = 120
Hence, Required number of ways = 210 x 120 = 25200

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Correct Answer : A

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Explanation
Taking all person of same nationality as one person, then we will have only three people.
These three person can be arranged themselves in 3! Ways.
8 Indians can be arranged themselves in 8! Way.
4 American can be arranged themselves in 4! Ways.
4 Englishman can be arranged themselves in 4! Ways.
Hence, required number of ways = 3!*8!*4!*4! Ways.

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Correct Answer : B

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Explanation
A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways.
Now, the captain can be selected from these 6 players in 6 ways.
Therefore, total ways the selection can be made is 210*6 = 1260.
Alternatively, we can select the 5 member team out of the 10 in 10C5 ways = 252 ways.
The captain can be selected from amongst the remaining 5 players in 5 ways.
Therefore, total ways the selection of 5 players and a captain can be made = 252*5 = 1260.

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Correct Answer : D

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Explanation

In a group of 6 boys and 4 girls, four children are to be selected such that
at least one boy should be there.
Hence we have 4 choices as given below
We can select 4 boys ------(Option 1).
Number of ways to this = ⁶C₄
We can select 3 boys and 1 girl ------(Option 2)
Number of ways to this = ⁶C₃ x ⁴C₁
We can select 2 boys and 2 girls ------(Option 3)
Number of ways to this = ⁶C₂ x ⁴C₂
We can select 1 boy and 3 girls ------(Option 4)
Number of ways to this = ⁶C₁ x ⁴C₃
Total number of ways
= (⁶C₄) + (⁶C₃ x ⁴C₁) + (⁶C₂ x ⁴C₂) + (⁶C₁ x ⁴C₃)
= (⁶C₂) + (⁶C₃ x ⁴C₁) + (⁶C₂ x ⁴C₂) + (⁶C₁ x ⁴C₁) [Applied the formula ⁿC_r = ⁿC_{(n - r)} ]

$=\left[\frac{6\times 5}{2\times 1}\right]+\left[\left(\frac{6\times 5\times 4}{3\times 2\times 1}\right)\times 4\right]+\left[\left(\frac{6\times 5}{2\times 1}\right)\left(\frac{4\times 3}{2\times 1}\right)\right]+\left[6\times 4\right]$

= 15 + 80 + 90 + 24
= 209

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Correct Answer : A

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Explanation
PIZZA = 5 letters.
But two letters ZZ is of same kind.
Thus, required permutations,
= 5!/2!
= 120/2
= 60.

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Correct Answer : A

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Explanation

From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.
Hence we have the following 3 choices
We can select 5 men ------(Option 1)
Number of ways to do this = ⁷C₅
We can select 4 men and 1 woman ------(Option 2)
Number of ways to do this = ⁷C₄ x ⁶C₁
We can select 3 men and 2 women ------(Option 3)
Number of ways to do this = ⁷C₃ x ⁶C₂
Total number of ways
= ⁷C₅ + [⁷C₄ x ⁶C₁] + [⁷C₃ x ⁶C₂]
= ⁷C₂ + [⁷C₃ x ⁶C₁] + [⁷C₃ x ⁶C₂] [Applied the formula ⁿC_r = ⁿC_{(n - r)} ]

$=\left[\frac{7\times 6}{2\times 1}\right]+\left[\left(\frac{7\times 6\times 5}{3\times 2\times 1}\right)\times 6\right]+\left[\left(\frac{7\times 6\times 5}{3\times 2\times 1}\right)\times \left(\frac{6\times 5}{2\times 1}\right)\right]$

= 21 + 210 + 525 = 756

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Correct Answer : B

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Explanation
No. of ways in which 10 paper can arranged is 10! Ways.
When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers.
These 9 papers can be arranged in 9! Ways.
And two papers can be arranged themselves in 2! Ways.
No. of arrangement when best and worst paper do not come together,
= 10!- 9!.2! = 9!(10-2) = 8.9!.

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