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Test: Permutation & Combination- 2 - CAT MCQ


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10 Questions MCQ Test - Test: Permutation & Combination- 2

Test: Permutation & Combination- 2 for CAT 2024 is part of CAT preparation. The Test: Permutation & Combination- 2 questions and answers have been prepared according to the CAT exam syllabus.The Test: Permutation & Combination- 2 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Permutation & Combination- 2 below.
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Test: Permutation & Combination- 2 - Question 1

MCQ (Multiple Choice Questions) with solution are available for Practice, which would help you prepare for Permutation and Combination under Quantitative Aptitude. You can practice these practice quizzes as per your speed and improvise the topic. The same topic is covered under various competitive examinations like - CAT, GMAT, Bank PO, SSC and other competitive examinations.

Q. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? 

Detailed Solution for Test: Permutation & Combination- 2 - Question 1


Test: Permutation & Combination- 2 - Question 2

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there? 

Detailed Solution for Test: Permutation & Combination- 2 - Question 2

The total number of ways to select 4 children out of 10 total children is (104)=10∗9∗8∗74∗3∗2∗1=210. However, because at least one boy must be in the group, we have to subtract 1 to account for the case where all four children are girls (There is (44)=1 such case). The answer is 210−1=209.

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Test: Permutation & Combination- 2 - Question 3

From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?

Detailed Solution for Test: Permutation & Combination- 2 - Question 3

Test: Permutation & Combination- 2 - Question 4

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together? 

Detailed Solution for Test: Permutation & Combination- 2 - Question 4

Vowels in the word "CORPORATION" are O,O,A,I,O
Lets make it as CRPRTN(OOAIO)
This has 7 lettes, where R is twice so value = 7!/2!
= 2520
Vowel O is 3 times, so vowels can be arranged = 5!/3!
= 20
Total number of words = 2520 * 20 = 50400

Test: Permutation & Combination- 2 - Question 5

The sum of all the possible numbers of 4 digits formed by digits 3, 5, 5, and 6 using each digit once is

Detailed Solution for Test: Permutation & Combination- 2 - Question 5

Total number of different numbers of 4 digits formed by using these digits only once = 4!/2! = 12.

The numbers formed whose unit digit is 3 = 3

The number formed whose unit digit is 5 = 6

The number formed whose unit digit is 6 = 3

(Same rule is applicable for 10th, 100th and 1000th units)

Sum of the total number of unit digits = 3×3+5×6 +6×3 = 57

Sum of the 10th digit number = 57×10

Sum of the 100th digit number = 57×100

Sum of the 1000th digit number = 57×1000

Sum of all the numbers formed = 57(1+10+100 + 1000) = 63327.

Test: Permutation & Combination- 2 - Question 6

India and Australia player one-day international cricket series until anyone team win 4 matches. No match ended in a  draw. In how many ways can the series be won?

Detailed Solution for Test: Permutation & Combination- 2 - Question 6

So, India can win the series in 1+4+10+20 = 35 ways.
Similarly, Australia can win the series in 35 ways.
Hence, the total number of ways in which the series can be won is 35+35=70 ways.

Test: Permutation & Combination- 2 - Question 7

What is the number of whole numbers formed on the screen of a calculator which can be recognised as numbers with (unique) correct digits when they are read inverted? The greatest number that can be formed on the screen of the calculator is 999999.

Detailed Solution for Test: Permutation & Combination- 2 - Question 7

The digits which can be recognised as unique digits when they are Inverted in a calculator are 0, 1, 2, 5, 6, and 9 Since the number cannot begin with zero all the numbers having 0 at units place should be discarded for otherwise when reed upside down the number yen begin with 3 we now :1St the different posibililities

Thus, the number of required numbers = 7 + 62 + 62 x 7 + .. + 62 x 7

Test: Permutation & Combination- 2 - Question 8

The number of ways in which 4 squares can be chosen at random on a chess board such that they lie on a diagonal line?

Detailed Solution for Test: Permutation & Combination- 2 - Question 8

Divide the chess board into two parts by a diagonal line 80 To select 4 squares lying on
diagonal line we have to choose the squares lying on P1, Q1, P2Q2, P3, Q3, P4Q4 and BD.
Number of squares on one side of BD =4C4 +5C4 +6C4 + 7C4 = 56. Thus total squares on either side of BO = 56 x 2 = 112.
On the diagonal BD, we will have 8C4, squares=70. So total squares = 112 . 70 = 182.
Similarly on the other side of the diagonal line AC we will have 182 squares.
Hence total number of squares = 2 x 182 = 364

Test: Permutation & Combination- 2 - Question 9

There are 4 letters and 4 addressed envelopes. If each letter is randomly placed in an envelope, then in how many ways can wrong choices be made? 

Detailed Solution for Test: Permutation & Combination- 2 - Question 9

4 letters can go into 4 envelopes in 4! Ways, of these onIy one choice is the correct one.

So. total number of wrong choices = 4! - 1

Test: Permutation & Combination- 2 - Question 10

How many words can be formed with the letters of the word 'PATALIPUTRA' without changing the relative order of the vowels and consonants?

Detailed Solution for Test: Permutation & Combination- 2 - Question 10

There are 11 letters in the word 'PATALIPUTRA‘ and there are two P's. two T's. three A's and four other different letters. Number of consonants = 6. number of vowels = 5.

Since relative order of the vowels and consonants remains unchanged. vowels can occupy only vowel’s place and consonants can occupy only consonants place.

Now 6 consonants can be arranged among themselves 6!/2!2! ways (since there are two P's and two T’s).

And 5 vowels can be arranged among themselves in 5!/3!ways (since A occurs thrice)

Therefore, Required number = 6!/2!2! x 5!/3! = 3600.

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