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Test:- Permutations And Combinations - 11


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20 Questions MCQ Test Topic-wise Tests & Solved Examples for IIT JAM Mathematics | Test:- Permutations And Combinations - 11

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Test:- Permutations And Combinations - 11 - Question 1

In a club election the number contestants is one more than the number of maximum candidates for which a voter can vote. If the total number of ways in which a voter can vote be 62 then the number of candidates is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 1

Test:- Permutations And Combinations - 11 - Question 2

The total number of selections of at most n things from (2n + 1) different things is 63. Then the value of n is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 2

Test:- Permutations And Combinations - 11 - Question 3

 Let 1 ≤ m < n ≤ p. The number of subsets of the set A - {1, 2, 3, p} having m, n as the least and the greatest elemens respectively, is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 3

Total number of subsets
= The number of selections of at least two elements inluding m, n and natural numbers lying between m and n.
= total number of selections fro n - m - 1 different things
= 2n-m-1

Test:- Permutations And Combinations - 11 - Question 4

The number of ways in which n different prizes can be distributed among m(< n) persons if each is entitled to receive at most n - 1 prizes, is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 4

Total number of ways = m x m x ... to n times = mn.

Test:- Permutations And Combinations - 11 - Question 5

The number of possible outcomes in a throw of n ordinary dice in which at least one of the dice shown an odd number is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 5

Total number of ways = 6 x 6 x ... to n times = 6n
Total number of ways to show only even numbers
= 3 x 3 x ... to n times = 3n.
∴ the required number of ways = 6n - 3n.

Test:- Permutations And Combinations - 11 - Question 6

The number of different 6 digit numbers that can be formed using the three digits 0, 1 and 2 is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 6


0 cannot go in the units place.
∴ the required number o f numbers = 2 * 35.

Test:- Permutations And Combinations - 11 - Question 7

The number of different matrices that can be formed with elements 0, 1, 2 or 3, each matrix having 4 elements, is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 7

The matrix will be of the order 4 M or 1 x 4 or 2 x 2.
For each order, the number of different coordinates nP3.
∴ the required number of triplets = n3-n(n-1)(n-2)

Test:- Permutations And Combinations - 11 - Question 8

Let A be a set of (n ≥ 3) distinct elements. The number of triplets (x, y, z) of the elements of .4 in which at. least two coordinates are equal is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 8

Total - none equal = n3-nP3

Test:- Permutations And Combinations - 11 - Question 9

The number of different pairs of words that can be made with the letters of the wrord STATICS is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 9

From every arrangement of 7 letters we get a pair by putting a sign after the four letters from the left.
∴ the required number of pairs = 7!/2!2!

Test:- Permutations And Combinations - 11 - Question 10

 Total number of 6-digit number in which all the odd digits and only odd digits appear, is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 10

Clearly, one of the odd digits 1, 3, 5, 7, 9 will be repreated.
The number of selections of the sixth digit = 5C1= 5.
∴ the required number ot numbers = 

Test:- Permutations And Combinations - 11 - Question 11

The number of divisors of the form 4n + 2(n ≥ 0) of the integer 240 is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 11


∴ the required number of divisors
= the number of selections of one 2 from four 2’s, any number of 3’s from one 3 and any number of 5’s from one 5.
= 1 x 2 x 2 = 4.

Test:- Permutations And Combinations - 11 - Question 12

In the next World Cup of cricket there will be 12 teams, divided equally in two groups. Teams of each group will play a match against each other. From each group 3 top teams will qualify for the next round. In this round, each team will play against others once. Four top teams of this round will qualify for the semifinal round, where each team will play against the others once. Two top teams of this round will go to the final round, where they will play the best of three matches. The minimum number of matches in the next World Cup will be

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 12

The number of marches in the first round = 6C2 + 6C2
The number of matches in the next round = 6C2
The number of matches in the semifinal round = 4C2.
∴the required number of matches

Note: For “best of three” at least two matches are played.

Test:- Permutations And Combinations - 11 - Question 13

The number of different ways in which 8 persons can stand in a row so that between two particular persons A and B there are always two persons, is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 13

The number of selection of 4 persons including A, B = 6C2.
Considering these four as a group, number of ar rangements with the other four = 5!
But in each group the number of arrangements = 2! x 2!
∴ the required number of ways =

Test:- Permutations And Combinations - 11 - Question 14

Four couples (husband and wife) decide to form a committee of four members. The number of different committees that can be formed in which no couple finds a place is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 14

The number of committees of 4 gentlemen =4C4 = 1
The number of committees of 3 gentlemen, 1 wife = 4C3 x 1C1
(∴ after selecting 3 gentlemen only 1 wife is left who can be included).
The number- of committees of 2 gentlemen, 2 wives
= 4C2 x 2C2
The number of committees of 1 gentlemen. 3 wives
= 4C1 x 3C3.
The number of committees of 4 wives = 1
∴ the required number of committes = 1 + 4 + 6 + 4+1 = 16

Test:- Permutations And Combinations - 11 - Question 15

From 4 gentlemen and 6 ladies a committee of five is to be selected. The number of ways in which the committee can be formed so that gentlemen are in majority is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 15

The committee will consist of 4 gentlemen and 1 lady or 3 gentlemen and 2 ladies.
∴ the number of committes = 

Test:- Permutations And Combinations - 11 - Question 16

There are 20 questions in a question paper. If no two students solve the same combination of questions but solve equal number of questions then the maximum number of students who appeared in the examination is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 16

If r questions are solved by each student then the number of possible selections of questions is 20Cr
∴ the number of students =20Cr
(∴ each student has solved different combination of questions)
∴ the maximum number of students = maximum value of 20Cr = 20C10, because 20C10 is the largest among 20C0, 20C1, ....20C20 - being the middle one. 

Test:- Permutations And Combinations - 11 - Question 17

Nine hundred distinct n-digit positive numbers are to be formed using only the digits 2, 5 and 7. The smallest value of n for which this is possible is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 17

Each place can be filled in 3 ways.
so, n places can be filled in 3 x 3 x ... to n times = 3n.
36= 729, 37=2187. So, for 3n-3 900 the smallest integral n is 7.

Test:- Permutations And Combinations - 11 - Question 18

The total number of integral solutions for (x, y, z) such that xyz = 24 is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 18

24 = 2 x 3 x 4 , 2 x 2 x 6 , 1 x 6 x 4, 1 x 3 x 8 , 1 x 2 x 12, 1 x 1 x 24
(as product of three positive integers)
∴ the total number of positive integral solutions of xyy. = 24 is equal to 

Any two of the factors in each factorization may be negative.
∴ the number of ways to associate negative sign in each case is 3C2, i.e., 3.
∴ the total number of integral solutions = 30 +3 x 3 = 120 

Test:- Permutations And Combinations - 11 - Question 19

The number of ways in which the- letters of the word ARTICLE can be rearranged so that the even places are always occupied by consonants is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 19

The number of ways to fill the three even places by 4 constonants = 4P3.
After filling the even places, remaining places can be filled in 4P4 ways.
So, the required number of words = 4P3 x 4P4.
 

Test:- Permutations And Combinations - 11 - Question 20

A cabinet of ministers consists of 11 ministers, one minister being the chief minister. A meeting is to be held in a room having a round table and 11 chairs round it, one of them being meant for the chairman. The number of ways in which the minister can take their chairs, the chief minister occupying the chairman's place, is

Detailed Solution for Test:- Permutations And Combinations - 11 - Question 20

pivot has been created by chairman’s place

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