Permutation And Combination - Aptitude MCQ Questions and Solutions with Explanations

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.

APPLE = 5 letters.
But two letters PP is of same kind.
Thus, required permutations,
$$\eqalign{ & = \frac{{5!}}{{2!}} \cr & = \frac{{120}}{2} \cr & = 60 \cr} $$

ALLAHABAD = 9 letters. Out of these 9 letters there is 4 A's and 2 L's are there.
So, permutations = $$\frac{{9!}}{{4!.2!}}$$ = 7560

(a) There are 4 vowels and all are alike i.e. 4A's.
_2^nd _4^th _6^th _8^th _
These even places can be occupied by 4 vowels. In $$\frac{{4!}}{{4!}}$$ = 1 Way.
In other five places 5 other letter can be occupied of which two are alike i.e. 2L's.
Number of ways = $$\frac{{5!}}{{2!}}$$ Ways.
Hence, total number of ways in which vowels occupy the even places = $$\frac{{5!}}{{2!}}$$ × 1 = 60 ways.

(b) Taking both L's together and treating them as one letter we have 8 letters out of which A repeats 4 times and others are distinct.
These 8 letters can be arranged in $$\frac{{8!}}{{4!}}$$ = 1680 ways.
Also two L can be arranged themselves in 2! ways.
So, Total no. of ways in which L are together = 1680 × 2 = 3360 ways.
Now,
Total arrangement in which L never occur together,
= Total arrangement - Total no. of ways in which L occur together.
= 7560 - 3360
= 4200 ways

The question asks for the number of ways to arrange 10 examination papers so that the best and the worst papers never come together.

This type of problem, where certain items "never come together", has a common strategy:
Total Arrangements - Arrangements Where They ARE Together = Arrangements Where They Are NEVER Together

Let's break it down step-by-step:

Step 1: Calculate the Total Number of Ways to Arrange all 10 Papers.
If we have 10 different papers, and we want to arrange them in a line, the number of ways is given by 10 factorial (10!).
10! means 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.
So, Total arrangements = 10!

Step 2: Calculate the Number of Ways where the Best and Worst Papers ARE Together.
To make sure the best and worst papers are always together, we can think of them as a single "block" or "unit".
Imagine tying the best paper and the worst paper together. Now, instead of 10 individual papers, you have:
(Paper 1), (Paper 2), ..., (Paper 8), and (Best & Worst Paper Block).
This means you now have 9 items to arrange (8 individual papers + 1 block).
These 9 items can be arranged in 9 factorial (9!) ways.

BUT, inside the "Best & Worst Paper Block", the best paper and the worst paper can swap their positions! They can be arranged in 2 factorial (2!) ways:
(Best, Worst) or (Worst, Best)

So, the total number of arrangements where the best and worst papers are together is: 9! × 2!

Step 3: Calculate the Number of Ways where the Best and Worst Papers NEVER Come Together.
Now we use our strategy:
Never Together = Total Arrangements - Arrangements Where They ARE Together
Never Together = 10! - (9! × 2!)

Step 4: Simplify the Expression.
We know that 10! can also be written as 10 × 9!
And 2! is 2 × 1 = 2.

So, our equation becomes:
Never Together = (10 × 9!) - (9! × 2)
You can see that 9! is common in both parts. Let's factor it out:
Never Together = 9! × (10 - 2)
Never Together = 9! × 8
Or, simply, 8 × 9!

This matches Option A.

Alternative way
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!

Let the Arrangement be,
B G B G B G B
4 boys can be seated in 4! Ways
Girl can be seated in 3! Ways
Required number of ways,
= 4! × 3!
= 144

⁵C₁ × ³C₁ - 1
= 15 - 1
= 14

Number of ways
$$\eqalign{ & { = ^{20}}{{\text{C}}_2} \cr & = \frac{{20!}}{{2! \times 18!}} \cr & = 20 \times \frac{{19}}{2} \cr & = 190 \cr} $$

There are 6 candidates and a voter has to vote for any two of them.
So, the required number of ways is,
$$\eqalign{ & { = ^6}{{\text{C}}_2} \cr & = \frac{{6!}}{{2! \times 4!}} \cr & = 15 \cr} $$

There 10 questions in part A out of which 8 question can be chosen as = ¹⁰C₈
Similarly, 5 questions can be chosen from 10 questions of Part B as = ¹⁰C₅
Hence, total number of ways,
$$\eqalign{ & { = ^{10}}{{\text{C}}_8}{ \times ^{10}}{{\text{C}}_5} \cr & = \frac{{10!}}{{2! \times 8!}} \times \frac{{10!}}{{5! \times 5}} \cr & = \left\{ {10 \times \frac{9}{2}} \right\} \times \left\{ {\frac{{10 \times 9 \times 8 \times 7 \times 6}}{{5 \times 4 \times 3 \times 2 \times 1}}} \right\} \cr & = 11340 \cr} $$

A triangle needs 3 points.
And polygon of 8 sides has 8 angular points.
Hence, number of triangle formed,
$$\eqalign{ & { = ^8}{{\text{C}}_3} \cr & = \frac{{8 \times 7 \times 6}}{{1 \times 2 \times 3}} \cr & = 56 \cr} $$