JEE Exam  >  JEE Tests  >  Daily Test for JEE Preparation  >  Maths Unit Test: Permutations & Combinations(June 28) - JEE MCQ

Maths Unit Test: Permutations & Combinations(June 28) - JEE MCQ


Test Description

15 Questions MCQ Test Daily Test for JEE Preparation - Maths Unit Test: Permutations & Combinations(June 28)

Maths Unit Test: Permutations & Combinations(June 28) for JEE 2024 is part of Daily Test for JEE Preparation preparation. The Maths Unit Test: Permutations & Combinations(June 28) questions and answers have been prepared according to the JEE exam syllabus.The Maths Unit Test: Permutations & Combinations(June 28) MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Maths Unit Test: Permutations & Combinations(June 28) below.
Solutions of Maths Unit Test: Permutations & Combinations(June 28) questions in English are available as part of our Daily Test for JEE Preparation for JEE & Maths Unit Test: Permutations & Combinations(June 28) solutions in Hindi for Daily Test for JEE Preparation course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Maths Unit Test: Permutations & Combinations(June 28) | 15 questions in 20 minutes | Mock test for JEE preparation | Free important questions MCQ to study Daily Test for JEE Preparation for JEE Exam | Download free PDF with solutions
*Answer can only contain numeric values
Maths Unit Test: Permutations & Combinations(June 28) - Question 1

A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then b + 3 g is equal to ____. (in integers)


Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 1

bC3 × gC2 = 168
b(b - 1)(b - 2) (g)(g - 1) = 8 × 7 × 6 × 3 × 2
By comparing both sides, we get
Number of boys = 8 and number of girls = 3
So, b + 3g = 17

*Answer can only contain numeric values
Maths Unit Test: Permutations & Combinations(June 28) - Question 2

The number of words (with or without meaning) that can be formed from all the letters of the word "LETTER" in which vowels never come together is ________. (in integers)


Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 2

LETTER
Vowels = EE, Consonants = LTTR
_L_T_T_R_
Total number of words = 4! / 2! × 5C2 × 2! / 2! = 12 × 10 = 120

1 Crore+ students have signed up on EduRev. Have you? Download the App
*Answer can only contain numeric values
Maths Unit Test: Permutations & Combinations(June 28) - Question 3

If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without meaning) be listed as in a dictionary, then the position of the word 'MOTHER' is _______. (in integers)


Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 3

EHMORT in alphabetical order:



Rank = 2 × 5! + 2 × 4! + 3 × 3! + 2! + 1
= 309

*Answer can only contain numeric values
Maths Unit Test: Permutations & Combinations(June 28) - Question 4

The number of six-letter words (with or without meaning), formed using all the letters of the word 'VOWELS', so that all the consonants never come together, is ______. (in integers)


Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 4


All consonants should not be together.
= Total - All consonants together
= 6! - 3! 4! = 576

*Answer can only contain numeric values
Maths Unit Test: Permutations & Combinations(June 28) - Question 5

If the digits are not allowed to repeat in any number formed by using the digits 0, 2, 4, 6, 8, then the number of all numbers greater than 10,000 is equal to _______. (in integers)


Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 5


 4 × 4 × 3 × 2 = 96

Maths Unit Test: Permutations & Combinations(June 28) - Question 6

If a, b and c are the greatest values of 19Cp, 20Cq and 21Cr respectively, then

Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 6

a = 19C10 or 19C9
b = 20C10
c = 21C10 or 21C11
1 = 
 

*Answer can only contain numeric values
Maths Unit Test: Permutations & Combinations(June 28) - Question 7

In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, -2 marks for each wrong answer and 0 marks if the question is not attempted. Then the number of ways a student appearing in the examination gets 5 marks is _____. (in integer)


Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 7

Let the student mark x correct answers and y incorrect.
So, 3x - 2y = 5 and x + y  5 where x, y  W
The only possible solution is (x, y) = (3, 2).
The student can mark the correct answer by only one choice, but for incorrect answer, there are two choices.
So, total number of ways of scoring 5 marks = 5C3(1)3 . (2)2 = 40

*Answer can only contain numeric values
Maths Unit Test: Permutations & Combinations(June 28) - Question 8

Let b1b2b3b4 be a 4-element permutation with bi  {1, 2, 3, ..., 100} for 1  i ≤ 4 and bi  bj for i ≠ j, such that either b1, b2, b3 are consecutive integers or b2, b3, b4 are consecutive integers.

Then the number of such permutations b1b2b3b4 is equal to _________. (in integers)


Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 8

Maths Unit Test: Permutations & Combinations(June 28) - Question 9

Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is

Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 9

Exactly 1 digit will repeat which can be selected in 5C1 ways.
 Total number of ways = 5C1.

*Answer can only contain numeric values
Maths Unit Test: Permutations & Combinations(June 28) - Question 10

The total number of three-digit numbers, with one digit repeated exactly two times, is ______. (in integers)


Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 10

If 0 is taken twice, then number of ways = 9
If 0 is taken once, then 9C1 × 2 = 18
If 0 is not taken, then 9C1 8C1.3 = 216
Total = 243

*Answer can only contain numeric values
Maths Unit Test: Permutations & Combinations(June 28) - Question 11

Let denote nCk and .
If Ak =  and A4 - A3 = 190 p, then p is equal to __________. (in integers)


Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 11

Ak = 
Ak = 21Ck + 21Ck = 2. 21Ck
A4 - A3 = 2 (21C4 - 21C3) = 2(5985 - 1330)
190p = 2(5985 - 1330)  p = 49

*Answer can only contain numeric values
Maths Unit Test: Permutations & Combinations(June 28) - Question 12

The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is _______. (in integers)


Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 12

First we arrange 5 red cubes in a row and assume there are x1, x2, x3, x4, x5 and x6 blue cubes between them.
Here, x1 + x2 + x3 + x+ x5 + x6 = 11 and x2, x3, x4, x≥ 2; x1,x6 ≥ 0
Let x2 = t+ 2, x3 = t3 + 2, x4 = t4 + 2, x5 = t5 + 2, where x1,t2, t3, t4, t5,x6 ≥ 0
So x1 + t2 + t3 + t4 + t5 + x6 = 3
Required number of solutions = 8C5 = 56

*Answer can only contain numeric values
Maths Unit Test: Permutations & Combinations(June 28) - Question 13

The number of 5-digit natural numbers, such that the product of their digits is 36, is _____ . (in integers)


Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 13

Factors of 36 = 22⋅ 32⋅ 1
Five-digit combinations can be
(1, 2, 2, 3, 3) (1, 4, 3, 3, 1), (1, 9, 2, 2, 1)
(1, 4, 9, 1, 1) (1, 2, 3, 6, 1) (1, 6, 6, 1, 1)
i.e., total numbers

= (30 × 3) + 20 + 60 + 10 = 180

*Answer can only contain numeric values
Maths Unit Test: Permutations & Combinations(June 28) - Question 14

An urn contains 5 red marbles, 4 black marbles and 3 white marbles. Then the number of ways in which 4 marbles can be drawn so that at the most three of them are red is _________. (in integers)


Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 14

Required number of ways = Total - (All selected marbles are red)
= 12C4 - 5C4
= 495 - 5 = 490

Maths Unit Test: Permutations & Combinations(June 28) - Question 15

Let n > 2 be an integer. Suppose that there are n Metro stations in a city located along a circular path. Each pair of stations is connected by a straight track only. Further, each pair of nearest stations is connected by blue line, whereas all remaining pairs of stations are connected by red line. If the number of red lines is 99 times the number of blue lines, then the value of n is:

Detailed Solution for Maths Unit Test: Permutations & Combinations(June 28) - Question 15

Number of two consecutive stations = n
Number of two non-consecutive stations = nc2 - n
Now, according to the question,
⇒ nc2 - n = 99n
 ⇒ - 100n = 0
⇒ n - 1 - 200 = 0
⇒ n = 201

360 tests
Information about Maths Unit Test: Permutations & Combinations(June 28) Page
In this test you can find the Exam questions for Maths Unit Test: Permutations & Combinations(June 28) solved & explained in the simplest way possible. Besides giving Questions and answers for Maths Unit Test: Permutations & Combinations(June 28), EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

Download as PDF

Top Courses for JEE