Permutation And Combination - Aptitude MCQ Questions and Solutions with Explanations

Total number of arrangements = $$\frac{{8!}}{{4! \times 2! \times 2!}}$$   = 420

Since, there are two 1's and two 0's, the number of arrangements in which the first 1 is before the first 2 is same as the number of arrangement in which the first 2 is before the first 1 and they are each equal to half the total number of arrangements = 210

If the two P’s were distinct (they could have different subscripts and colours), the number of possible permutations would have been 5! = 120

For example let us consider one permutation: P₁LEAP₂

Now if we permute the P’s amongst them we still get the same word PLEAP. The two P’s can be permuted amongst them in 2! ways.

We were counting P₁LEAP₂ and P₂LEAP₁ as different arrangements only because we were artificially distinguishing between the two P’s Hence the number of different five letter words that can be formed is:
= $$\frac{{5!}}{{\left(1!\right) \left(2!\right) \left(1!\right) \left(1!\right)}}$$
= 60

The situation is similar to placing 10 identical balls among 5 distinguishable boxes, where a box may have zero or more balls in it.

This case can be represented as arranging ten balls and (5 - 1) four walls in the single row, which can be done in ¹⁴C₄ ways. (The balls placed between every successive pair of walls belong to one group) ¹⁴C₄ = 1001 ways.

¹⁰C₅ : when both are not included.
¹⁰C₄ : when one of them is included.

Number of ways
= ¹⁰C₅ + ¹⁰C₄ + ¹⁰C₄
= 672

There are 9 × 10⁴ = 90000, 5-digit positive integers.

Out of these 90000 positive integers, the sum of the digits of half of the numbers will add up to an odd number and the remaining half will add up to an even number.

Hence, there are $$\frac{{90000}}{2}$$  = 45000, 5-digit positive integers whose sum add up to an odd number.

All the boxes contain distinct number of chocolates.
For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on.

The number of ways of giving 4 boxes to the 4 person is,
⁸C₄ = 70

First let us consider the 28 unoccupied seats.
They create 29 slots - one on the left of each seat and one on the right of the last one.
We can place the 7 friends in any of these 29 slots i.e. ²⁹P₇ ways.

Each of the 10 letters can be posted in any of the 5 boxes.
So, the first letter has 5 options, so does the second letter and so on and so forth for all of the 10 letters.
i.e.5 × 5 × 5 × . . . . × 5 (up-to 10 times).
= 5¹⁰

Let there were x teams participating in the games, then total number of matches,
ⁿC₂ = 153
On solving we get,
⇒ n = −17 and n =18
It cannot be negative so,
n = 18 is the answer.

Six balls can be selected in the following ways:
One red ball and 5 blue ball
Or
Two red balls and 4 blue balls.
Total number of ways:
= (³C₁ × ⁷C₅) + (³C₂ × ⁷C₄)
= 63 + 105
= 168