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Test:- Permutations And Combinations - 10 - Mathematics MCQ


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20 Questions MCQ Test Topic-wise Tests & Solved Examples for Mathematics - Test:- Permutations And Combinations - 10

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Test:- Permutations And Combinations - 10 - Question 1

The number of 4-digit numbers that can be made with the digits 1,2, 3, 4 and 5 in which at least two digits are identical, is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 1

The number of numbers when repelition is allowed = 54.
The number of numbers when digits cannot be repeated = 5P5.
So, the required number of numbers = 54-5!

Test:- Permutations And Combinations - 10 - Question 2

The number of words that can be made by rearranging the letters of the word APURBA so that vowels and consonants alternate is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 2

First start with consonants and then with vowels.

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Test:- Permutations And Combinations - 10 - Question 3

The number of words that can be made by writing down the letters of the word CALCULATE such that each word starts and ends with a consonant, is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 3

The number of words like C-C=   the number of words like L - L.
The number of words beginning or ending with

The number of words beginning or ending with (C or L).

∴ the required number of words

Test:- Permutations And Combinations - 10 - Question 4

The number of numbers of 9 different non-zero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 4


The required number of numbers = 4P4 x 4P4

Test:- Permutations And Combinations - 10 - Question 5

In the decimal system of numeration the number of 6-digit numbers in which the digit in any place is greater than the- digit to the left of it is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 5

Any selection of six digits from 9 digits (excluding 0) will give a number of the required variety.
∴ the required number of numbers =9C6

Test:- Permutations And Combinations - 10 - Question 6

The number of 5 digit numbers in which no two consecutive digits are identical is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 6

x x x x x
Ways 9 9 9 9 9
0 cannot be placed in the first place In the next place any digit except the one used in the first place can be used, etc.

Test:- Permutations And Combinations - 10 - Question 7

The number of 6-digit numbers in which the sum of the digits is divisible by 5 is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 7

x x x x x x
Wavs 9 10 10 10 10 6 
0 cannot be filled in the first place. In other four places any digit can be filled. After filling the first five places, the last place can be filled by 0 or 5, 1 or 6, 2 or 7, 3 or 8, 4 or 9 depending upon whether the sum or five digits filled is of the form 5m, 5m + 4, 5m + 3, 5m + 2 or 5m + 1 respectively.
∴ The required number of numbers = 9 x 104 x 2

Test:- Permutations And Combinations - 10 - Question 8

The sum of all the numbers of four different digits that can be made by using the digits 0, 1, 2 and 3 is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 8

The number of numbers with 0 in the units place = 3! = 6
The number of numbers with 1 or 2 or 3 in the units place - 3! - 2! = 4
∴ the sum of the digit s in the units place = 6 x 0 +4 x 1 +4 x 2 +4 x 3= 24
Similarly for the tens and the hundreds places.
The number of numbers with 1 or 2 or 3 in the thousands place = 3!
∴ the sum of the digits in the thousands place = 6 x1 + 6 x 2 + 6 x 3 =36
∴ the required sum = 36 x 1000 + 24 x 10 + 24

Test:- Permutations And Combinations - 10 - Question 9

A teacher takes 3 children from her class to the zoo at a time as often as she can, but she does not lake the same three children to the zoo more than once. She finds that she goes to the zoo 84 times more than a particular child goes to the zoo. The number of children in her class is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 9

The number of times the teacher goes to the zoo = nC3.
The number of times a particular child goes to the zoo
= n-1C2.
From the question nC3 - n-1C2 = 84
or
(n - 1) ( n - 2) (n - 3) = 6 x 84 = 9 x 8 x 7 
⇒ n - 1 = 9

Test:- Permutations And Combinations - 10 - Question 10

 ABCD is a convex quadrilateral, 3, 4, 5 and 6 points are marked on the sides AB, BC, CD and DA respectively. The number of triangles with vertices on different sides is

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

The number of triangles with vertices on sides
AB, BC, CD = 3C1 x 4C1 x 5C1
Similarly for other cases.

∴ the total number of triangles

Test:- Permutations And Combinations - 10 - Question 11

There are 10 points in a plane of which no three points are collinear and 4 points are concyclic. The number of different circles that can be drawn through at least 3 points of these points is

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

The number of circles = (10C3-4C3) + 1

Test:- Permutations And Combinations - 10 - Question 12

In a polygon the number of diagonals is 54. The number of sides of the polygon is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 12

n(n -3) /2 = 54 ⇒ n = 12

Test:- Permutations And Combinations - 10 - Question 13

In a polygon no three diagonals are concurrent, if the total number of points of intersection of diagonals interior to the polygon be 70 then the number of diagonals of the polygon is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 13

A selection of four vertices of the polygon gives an interior intersection.
∴ the number of sides = n 
nC4 = 70

⇒ n(n-1)(n-2)(n-3) = 24 x 70 = 8 x 7 x 6 x 5
⇒ n = 8
∴ the number of diagonals =8C2- 8

Test:- Permutations And Combinations - 10 - Question 14

n lines are drawn in a plane such that no two of them are parallel and no three of them are concurrent. The number of different points at which these lines will cut is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 14

The number of ways of selecting 2 lines from n lines nC2

Test:- Permutations And Combinations - 10 - Question 15

The number of triangles that can be formed with 10 points as vertices, n of them being collinear, is 110. Then n is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 15

Test:- Permutations And Combinations - 10 - Question 16

There are three coplanar parallel lines. If any p points are taken on each of the lines, the maximum number of triangles with vertices at these points is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 16

The number of triangles with vertices on one line and the third vertex on any one of the other two lines.

∴ the required number of triangles = p3 +3p2(p-1)
Note: The word “maximum” ensures that no selection of points from each of the three lines are collinear.

Test:- Permutations And Combinations - 10 - Question 17

Two teams are to play a series of 5 matches between them. A match ends in a win or loss or draw for a term. A number of people forecast the result of each match and no two people make the same forecast for the series of matches. The smallest group of people in which one person forecasts correctly for all the matches will contain n people, where n is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 17

The smallest number of people
= total number of possible forecasts
= total number of possible results
= 3 * 3 * 3 * 3 * 3

Test:- Permutations And Combinations - 10 - Question 18

A bag contains 3 black, 4 white and 2 red balls, all the balls being different. The number of selections of at most 6 balls cotaining balls of all the colours is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 18

The required number of selections.

Test:- Permutations And Combinations - 10 - Question 19

In an examination of 9 papers a candidate has to pass in more papers than the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 19

The candidate is unsuccessful if he fails in 9 or 8 or 7 or 6 or 5 papers.
∴ the number of ways to be unsuccessful

Test:- Permutations And Combinations - 10 - Question 20

The number of 5-digit numbers that can be made using the digits 1 and 2 and in which at least one digit is different, is

Detailed Solution for Test:- Permutations And Combinations - 10 - Question 20

Total number of numbers without restriction 25.
Two numbers have all the digits equal. So, the required number of numbers = 25 - 2

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