# Permutation and Combination Questions for XAT Set-3 PDF

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### Permutation and Combination Questions for XAT Set-3 PDF

Download important Permutation and Combination Questions Set-3 for XAT  Set-2 PDF based on previously asked questions in the XAT exam. Practice Permutation and Combination Questions  Set-3 PDF for the XAT exam.

Question 1: If all letters of the word “CHCJL” be arranged in an English dictionary, what will be the $50^{th}$ word?

a) HCCLJ

b) LCCHJ

c) LCCJH

d) JHCLC

e) None of the above

Question 2: The coefficient of $x^{7}$ in the expansion of $(1 – x^{2} + x^{3})(1 + x)^{10}$ is:

a) 75

b) 78

c) 85

d) None of the above

Question 3: A doctor has decided to prescribe two new drugs D1and D2 to 200 heart patients such that 50 get drug D1, 50 get drug D2 and 100 get both. The 200 patients are chosen so that each had 80% chance of having a heart attack if given neither of the drugs. Drug D1 reduces the probability of a heart attack by 35 %, while drug D2 reduces the probability by 20%. The two drugs when taken together, work independently. If a patient, selected randomly from the chosen 200 patients, has a heart attack then the probability that the selected patient was given both the drug is:

a) 0.42

b) 0.49

c) 0.56

d) 0.40

Question 4: How many positive integers ‘n’ can we form using the digits 3, 4, 4, 5, 6, 6, 7 if we want ‘n’ to exceed 6,000,000?

a) 320

b) 360

c) 540

d) 720

Question 5: A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is:

a) 220

b) 600

c) 240

d) None of the above

Question 6: Out of 8 consonants and 5 vowels, how many words can be made, each containing 4 consonants and 3 vowels?

a) 700

b) 504000

c) 3528000

d) 7056000

Question 7: In a True/False quiz, 4 marks are awarded for each correct answer and 1 mark is deducted for each wrong answer. Amit, Benn and Chitra answered the same 10 questions, and their answers are given below in the same sequential order.
AMIT     T T F F T T F T T F
BENN    T T T F F T F T T F
CHITRA T T T T F F T F T T
If Amit and Benn both score 35 marks each then Chitra’s score will be:

a) 10

b) 15

c) 20

d) 25

e) None of the above

Question 8: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at 0. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and 0 so as to form a triangle?

Question 9: The numbers 1, 2, …, 9 are arranged in a 3 X 3 square grid in such a way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value.
If the top left and the top right entries of the grid are 6 and 2, respectively, then the bottom middle entry is

Question 10: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?

The alphabetical order = CCHJL

Number of words starting with C = $4! = 24$

Number of words starting with H = $\frac{4 !}{2} = 12$

Number of words starting with J = $\frac{4 !}{2} = 12$

Total words till now = 24 + 12 + 12 = 48

First word starting with L (49th in dictionary) = LCCHJ

Therefore, the 50th word =  LCCJH

We need to find the coefficient of $x^7$ in the expansion $(1-x^{2}+x^{3})(1 + x)^{10}$

Now, $(1 + x)^{10}$ will have all the powers of x from 0 to 10. Multiplying these powers by 1, $x^2$ and $x^3$ will yield different results but we are interested in finding only the coefficient of $x^7$. When we multiply $x^7$ of $(1 + x)^{10}$ by 1, $x^5$ of $(1 + x)^{10}$ by $x^2$ and  $x^4$ of $(1 + x)^{10}$ by $x^3$ we will get $x^7$. coefficient of $x^7$ in $(1 + x)^{10}$ is $10C_7$ = 120, coefficient of $x^5$ in $(1 + x)^{10}$ is $10C_5$ = 252, coefficient of $x^4$ in $(1 + x)^{10}$ is $10C_4$ = 210 adding 120 and 210 and subtracting(since $x^2$ has a negative sign) 252 we get coefficient of $x^7$ as 78

Therefore our answer is option ‘B’

Given that probability of getting a heart attack before any drug = 0.80

Drug D1 reduces the probability of a heart attack by 35 %. Therefore, the probability of a patient getting heart attack after he has taken D1 = (1-0.35)*0.80 = 0.52

Drug D2 reduces the probability of a heart attack by 20 %. Therefore, the probability of a patient getting heart attack after he has taken D2 = (1-0.20)*0.80 = 0.64

It is given that both D1 and D2 work independently. Therefore, the probability of a patient getting heart attack after he has taken both D1 and D2 = (1-0.35)*(1-0.20)*0.80 = 0.416

A total of 100 patients have taken both the drugs whereas only 50-50 patients took drug D1 and D2.

Hence, the probability that the selected patient was given both the drug is = $\dfrac{0.416}{0.416+0.5*0.52+0.5*0.64}$ = 0.417 $\approx$ 0.42

Therefore, we can say that option A is the correct answer.

We are given exactly 7 digits – 3, 4, 4, 5, 6, 6, 7. The millions digit can be either 6 or 7.

Case 1: When the millions digit is 6.

6 _ _ _ _ _ _ We are left with six digits {3, 4, 4, 5, 6, 7}.

These six digits can be arranged in six places in $\dfrac{6!}{2!}$ ways.

Case 1: When the millions digit is 7.

7 _ _ _ _ _ _ We are left with six digits {3, 4, 4, 5, 6, 6}.

These six digits can be arranged in six places in $\dfrac{6!}{2!*2!}$ ways.

Therefore, total number of numbers ‘n’ = $\dfrac{6!}{2!}$ + $\dfrac{6!}{2!*2!}$ = 360 + 180 = 540. Hence, option C is the correct answer.

For a number to be divisible by 3, the sum of it’s digit should be a multiple of 3. This can be done in two ways. When the number are formed by using either {1, 2, 3, 4, 5} or {0, 1, 2, 4, 5}.

Total 5 digit numbers that can be formed by {1, 2, 3, 4, 5} without repetition = 5*4*3*2*1 = 120

Total 5 digit numbers that can be formed by {0, 1, 2, 4, 5} without repetition = 4*4*3*2*1 = 96

Therefore, total such number = 120 + 96 = 216. Hence, option D is the correct answer.

Out of 8 consonants and 5 vowels selecting 4 consonants and 3 vowels can be done in $^8C_4 * ^5C_3 = \dfrac{8*7*6*5*5*4}{4*3*2*2}$
= 700
Internal arrangement of 7 letters = 7!
Thus, the total number of ways = 7!*700 = 352800
Hence, option C is the correct answer.

Both Amit and Ben scored 35 marks. 4 marks are awarded for a correct answer and 1 mark is deducted for an incorrect answer.
Let the number of questions that Amit got right be ‘x’.
=> 4x -(10-x) = 35
5x = 45
x = 9.
Therefore, Amit must have made only 1 mistake. The same must have been the case with Ben too.
The responses given by the 3 persons are as follows:

AMIT     T T F F T T F T T F
BENN    T T T F F T F T T F
CHITRA T T T T F F T F T T

Amit and Ben have given different responses for question 3 and question 5. Therefore, one of them must be wrong in each of these questions. Also, Amit must have given the correct answer for one of these 2 questions and Ben must have answered the other one correct, since both of them got 9 questions correct.

Let us assume Amit has given an incorrect response for question 3 and Ben has given an incorrect response for question 5.
In this case, Chitra’s would have given 4 correct responses (questions 1,2,3, and 9). Chitra’s score would have been 4*4 – 6 = 10.

Let us assume Amit has given an incorrect response for question 5 and Ben has given an incorrect response for question 3.
In this case, Chitra’s would have given 4 correct responses (questions 1,2,5, and 9). Chitra’s score would have been 4*4 – 6 = 10.

Therefore, Chitra’s score should have been 10 and hence, option A is the right answer.

The total number of given points are 11. (10 on circumference and 1 is the center)
So total possible triangles = 11C3 = 165.
However, AOB, COD, EOF, GOH, JOK lie on a straight line. Hence, these 5 triangles are not possible. Thus, the required number of triangles = 165 – 5 = 160

According to the question each column, each row, and each of the two diagonals of the 3X3 matrix add up to the same value. This value must be 15.

Let us consider the matrix as shown below:

Now we’ll try substituting values from 1 to 9 in the exact middle grid shown as ‘x’.

If x = 1 or 3, then the value in the left bottom grid will be more than 9 which is not possible.

x cannot be equal to 2.

If x = 4, value in the left bottom grid will be 9. But then addition of first column will come out to be more than 15. Hence, not possible.

If x=5, we get the grid as shown below:

Hence, for x = 5 all conditions are satisfied. We see that the bottom middle entry is 3.

Hence, 3 is the correct answer.

For the number to be divisible by 6, the sum of the digits should be divisible by 3 and the units digit should be even. Hence we have the digits as

Case I: 2, 3, 4, 6
Now the units place can be filled in three ways (2,4,6), and the remaining three places can be filled in 3! = 6 ways.
Hence total number of ways = 3*6 = 18

Case II: 0, 2, 3, 4
case II a: 0 is in the units place => 3! = 6 ways
case II b: 0 is not in the units place => units place can be filled in 2 ways( 2,4) , thousands place can be filled in 2 ways ( remaining 3 – 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 2 * 2 * 2 = 8
Total number of ways in this case = 6 + 8 = 14 ways.

Case III: 0, 2, 4, 6
case III a: 0 is in the units place => 3! = 6 ways

case II b: 0 is not in the units place => units place can be filled in 3 ways( 2,4,6) , thousands place can be filled in 2 ways (remaining 3 – 0) and remaining can be filled in 2! = 2 ways. Hence total number of ways = 3 * 2 * 2 = 12
Total number of ways in this case = 6 + 12= 18 ways.

Hence the total number of ways = 18 + 14  + 18 = 50 ways

We hope this Permutation and Combination Questions Set-3 PDF for XAT with Solutions will be helpful to you.