Test Level 2: Permutation & Combination - 2 - CAT MCQ

10 persons can be arranged in 10! ways.
if 1 person is placed in a room for single person = 10!/1!
If 2 persons are placed in room for double person, then the arrangements = 10!/2!
If 3 persons are placed in the room for three persons then number of arrangements = 10!/3!
If 4 persons are placed in a room for four people, then number of arrangements = 10!/4!
So, total number of arrangements = 10!/2!3!4!

Number of ways in which 'n' objects can be placed on 'n' positions in such a manner that none of them is correct is given by the dearrangement formula.
Dearrangement (n) = n!(1/0! - 1/1! + 1/2! - 1/3!.... 1/n!)
Number of ways 

Total number of five-digit numbers formed by the digits 1, 2, 3, 4 and 5 is 5!.
Therefore, exhaustive number of cases = 5! = 120
A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
Therefore, the last two digits can be 12, 24, 32 or 52. The last two digits can be filled in 4 ways.
Corresponding to each of these ways, there are 3! = 6 ways of filling the remaining three places.
∴ The total number of five-digit numbers formed by the digits 1, 2, 3, 4 and 5 and divisible by 4 is 4 × 6 = 24
Therefore, favourable number of cases = 24
So, required probability = 24/120 = 1/5

We require that the 1st and the 5th numbers are the same and the 2nd and the 4th numbers are the same. If the 1st and the 2nd numbers are chosen, the 5th and the 4th numbers will also have been chosen. As the 1st number cannot be 0, it can be chosen in 9 ways. The 2nd and the 3rd numbers can be chosen in 10 ways each. Therefore, the total number of ways is (9 × 10²) = 900.

4 men can be chosen from among 6 in ⁶C₄ ways and 3 women from among 4 can be chosen in ⁴C₃ ways, i.e. 60 ways.
If 2 women, W1 and W2, are in the same committee, then the 3rd woman can be chosen in 2 ways.
In this case, the number of ways will be ⁶C₄ × 2 ways.
Therefore, the number of ways of forming the committee if W1 and W2 are never in the same committee = 60 - 30 = 30

The code is formed by 4 letters and 4 digits. The 4 letters are first to be selected and then to be arranged, and the same has to be done for digits.
Number of ways to select and arrange the letters = ⁸C₄ × 4!
Number of ways to select and arrange the digits = ¹⁰C₄ × 4!
The code is formed by both letters and digits, so total number of cases = (⁸C₄ × 4!) × (¹⁰C₄ × 4!)

Ice-creams with no topping = 6
Ice creams with one topping = 6 × 5 = 30
Ice-creams with two toppings = 6 × ⁵C₂ = 60
Ice-creams with three toppings = 6 × ⁵C₃ = 60
Ice-creams with four toppings = 6 × ⁵C₄ = 30
Ice-creams with five toppings = 6 × ⁵C₅ = 6
So, total number of different ice-creams possible = 6 + 30 + 60 + 60 + 30 + 6 = 192 = 6 × 2⁵

For n > 3, take n = 4.

The total number of ways in which five tourists can stay in eight hotels is (8)⁵. The number of ways in which no two tourists stay together = 8 × 7 × 6 × 5 × 4 = 6720. Therefore, the number of ways in which at least two tourists stay together = (8)⁵ – 6720 = 32,768 - 6720 = 26,048

Since we require that there is exactly one 3, this digit could occupy any of the five places. If the number does not start with 3, the first digit must necessarily be a digit other than 3 and 0. In such a case, the first digit can be chosen in eight ways and the remaining digits can be chosen in 9 ways each. The numbers will be of the following forms: 3XXXX, X3XXX, XX3XX, XXX3X and XXXX3. Correspondingly, the total number of ways is (9⁴) + (4 × 8 × 9³) = 29889.

Let the number of green balls be x.
Then, the number of red balls is 2x. Let the number of blue balls be y.
Then, x + 2x + y = 10
⇒ 3x + y = 10
⇒ y = 10 - 3x
Clearly, x can take values 0, 1, 2 and 3. The corresponding values of y are 10, 7, 4 and 1.
Thus, the possibilities are (0, 10, 0), (2, 7, 1), (4, 4, 2) and (6, 1, 3), where (r, b, g) denotes the numbers of red, blue and green balls.
So, the answer is Option (2).

For the first 4 prizes, four students have to be selected. Number of ways to do this = ⁵⁰C₄.
Further, 3 prizes can be given to anyone. So, the possibility for each prize is 50.
Therefore, total number of possibilities = ⁵⁰C₄ × (50)³

Letters will occur in the order A, D, M, N, O and R.
The number of words which begin with A are ⁵P₅ = 120.
Similarly, there are 120 words which begin with D, 120 words which begin with M, 120 words which begin with N and 120 words which begin with O.
So, 601st word begins with R.
There are ³P₃ = 6 words which have RAD as the first three letters and ³P₃ = 6 words which have RAM as the first three letters.
So, after 600 + 6 + 6 = 612 words, we have the words which have RAN as first three letters.
613rd word is RANDMO and 614th word is RANDOM.
So, the answer is option (2).

The first and the last letters can be filled in five ways each.
The remaining letters can be filled in 26 ways each.
Therefore, the total number of ways in which a four-letter word can be formed is 5 × 26 × 26 × 5 = 16,900.

White can be selected in 32 ways (because in a chessboard, there are 32 white and 32 black squares). There are 4 black squares in any row and 4 in any column. So, black can be selected in (32 - 8) = 24 ways.
Thus, total ways of selecting 1 white and 1 black = 32 x 24 = 768

If 'n' things are there out of which 'r' has to follow an order, then the total number of arrangements = n!/r!
So, different ways in which Mr. Bhima can eat all the sweets, such that he has to eat A before B and B before C = 10!/3! = 604,800

4-lettered palindrome means the first and the last, and the 2^nd and the 3^rd letters should be the same.
The first letters can be any out of 26.
Similarly, the second letters also can be any out of 26.
So, the required number of palindromes is 26 × 26 = 262

Ten members in the committee means 1 member each from 10 different families.
The number of ways to select 10 families from 20 families is ²⁰C₁₀.
Now, for each family, there are two different possibilities for the selection of members. Therefore, the number of ways in which this can be done for the 10 families is 2¹⁰.
These two conditions co-exist. So, total number of cases = ²⁰C₁₀ × 2¹⁰

If the number is divisible by 6, it has to be divisible by both 2 and 3.
If it has to be divisible by 2, it has to be even.
If it is divisible by 3, then the sum of the digits has to be divisible by 3.
Possible 4-digit combinations are: (1, 2, 3, 6); (1, 2, 4, 5); (2, 3, 4, 6); (1, 3, 5, 6); (3, 4, 5, 6)
Number of cases for (1, 2, 3, 6) = 3 × 2 × 1 × 2 = 12
Number of cases for (1, 2, 4, 5) = 3 × 2 × 1 × 2 = 12
Number of cases for (2, 3, 4, 6) = 3 × 2 × 1 × 3 = 18
Number of cases for (1, 3, 5, 6) = 3 × 2 × 1 × 1 = 6
Number of cases for (3, 4, 5, 6) = 3 × 2 × 1 × 2 = 12
Total number of cases = 12 + 12 + 18 + 6 + 12 = 60

First digit should be 7 or 8 as the number should be greater than
7,00,000, so number of ways arranging that is 2.
Ways of arranging rest of the 5 digits will be 5!.
So, required number of natural numbers = 2 × 5! = 240