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Permutations And Combinations - MCQ 1 - Banking Exams MCQ


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20 Questions MCQ Test Quantitative Aptitude for Competitive Examinations - Permutations And Combinations - MCQ 1

Permutations And Combinations - MCQ 1 for Banking Exams 2024 is part of Quantitative Aptitude for Competitive Examinations preparation. The Permutations And Combinations - MCQ 1 questions and answers have been prepared according to the Banking Exams exam syllabus.The Permutations And Combinations - MCQ 1 MCQs are made for Banking Exams 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Permutations And Combinations - MCQ 1 below.
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Permutations And Combinations - MCQ 1 - Question 1

 How many 3 digit number can be formed with the digits 5, 6, 2, 3, 7 and 9 which are divisible by 5 and none of its digit is repeated?

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 1

Answer – c) 20 Explanation : _ _ 5
first two places can be filled in 5 and 4 ways respectively so, total number of 3 digit number = 5*4*1 = 20

Permutations And Combinations - MCQ 1 - Question 2

 In how many different ways can the letter of the word ELEPHANT be arranged so that vowels always occur together?

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 2

Answer – b) 2160 Explanation : Vowels = E, E and A. They can be arranged in 3!/2! Ways so total ways = 6!*(3!/2!) = 2160

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Permutations And Combinations - MCQ 1 - Question 3

A fruit basket contains 4 oranges, 5 apples and 6 mangoes. The number of ways a person make selection of fruits from among the fruits in the basket is?

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 3

Whenever it is not explicity mentioned that fruit and distint, We take them as identical. 0 or more orange can be selected from 4 identical oranges in (4+1)=5 ways. 0 or more apples can be selected from 5 identical apples in (5+1)=6 ways.

0 or more mangoes can be selected from 6 identical mangoes in 7 ways.

∴ Total no. of ways in which all of three types of fruits can be selected (the no. of any type of fruits may also be 0),

5×6×7=210,

But in these 20 selection, there is one selection where all fruits are 0, hence we reduce 1 selection.

∴ the required no. =210−1=209

Permutations And Combinations - MCQ 1 - Question 4

There are 15 points in a plane out of which 6 are collinear. Find the number of lines that can be formed from 15 points.

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 4

Answer – c) 91 Explanation : From 15 points number of lines formed = 15c2 6 points are collinear, number of lines formed by these = 6c2 So total lines = 15c2 – 6c2 + 1 = 91

Permutations And Combinations - MCQ 1 - Question 5

In how many ways 4 Indians, 5 Africans and 7 Japanese be seated in a row so that all person of same nationality sits together

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 5

Answer – a) 4! 5! 7! 3!
Explanation : 4 Indians can be seated together in 4! Ways, similarly for Africans and Japanese in 5! and 7! respectively. So total ways = 4! 5! 7! 3!

Permutations And Combinations - MCQ 1 - Question 6

In how many ways 5 Americans and 5 Indians be seated along a circular table, so that they are seated in alternative positions

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 6

Answer – c) 4! 5!
Explanation : First Indians can be seated along the circular table in 4! Ways and now Americans can be seated in 5! Ways. So 4! 5! Ways

Permutations And Combinations - MCQ 1 - Question 7

4 matches are to be played in a chess tournament. In how many ways can result be decided?

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 7

Answer – c) 81 Explanation : Every chess match can have three result i.e. win, loss and draw so now of ways = 3*3*3*3 = 81 ways

Permutations And Combinations - MCQ 1 - Question 8

There are 6 players in a cricket which is to be sent to Australian tour. The total number of members is 12.

Q. If 2 particular member is always included

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 8

Answer – a) 210 Explanation : only 4 players to select, so it can be done in 10c4 = 210

Permutations And Combinations - MCQ 1 - Question 9

There are 6 players in a cricket which is to be sent to Australian tour. The total number of members is 12.

Q. If 3 particular player is always excluded

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 9

Answer – c) 84 Explanation : 6 players to be selected from remaining 9 players in 9c6 = 84 ways

Permutations And Combinations - MCQ 1 - Question 10

In a group of 6 boys and 5 girls, 5 students have to be selected. In how many ways it can be done so that at least 2 boys are included

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 10

Answer – b) 426 Explanation : 6c2*5c3 + 6c3*5c2 + 6c4*5c1 + 6c5

Permutations And Combinations - MCQ 1 - Question 11

In how many ways can 5 boys and 4 girls can be seated in a row so that they are in alternate position.

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 11

Answer – b) 2880 Explanation : First boys are seated in 5 position in 5! Ways, now remaining 4 places can be filled by 4 girls in 4! Ways, so number of ways = 5! 4! = 2880

Permutations And Combinations - MCQ 1 - Question 12

In how many ways 5 African and five Indian can be seated along a circular table, so that they occupy alternate position.

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 12

Answer – b) 4! 5!
Explanation : First 5 African are seated along the circular table in (5-1)! Ways = 4!. Now Indian can be seated in 5! Ways, so 4! 5!

Permutations And Combinations - MCQ 1 - Question 13

There is meeting of 20 delegates is to be held in a hotel. In how many ways these delegates can be seated along a round table, if three particular delegates always seat together.

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 13

Answer – a) 17! 3!
Explanation : Total 20 persons, 3 always seat together, 17 + 1 =18 delegates can be seated in (18 -1)! Ways = 17! And now that three can be arranged in 3! Ways. So, 17! 3!

Permutations And Combinations - MCQ 1 - Question 14

In how many 8 prizes can be given to 3 boys, if all boys are equally eligible of getting the prize.

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 14

Answer – a) 512 Explanation : Prizes cab be given in 8*8*8 ways = 512 ways

Permutations And Combinations - MCQ 1 - Question 15

There are 15 points in a plane out of which 6 are collinear. Find the number of lines that can be formed from 15 points.

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 15

Answer – c) 91 Explanation : From 15 points number of lines formed = 15c2 6 points are collinear, number of lines formed by these = 6c2 So total lines = 15c2 – 6c2 + 1 = 91

Permutations And Combinations - MCQ 1 - Question 16

In party there is a total of 120 handshakes. If all the persons shakes hand with every other person. Then find the number of person present in the party.

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 16

Answer – b) 16 Explanation : Nc2 = 120 (N is the number of persons)

Permutations And Combinations - MCQ 1 - Question 17

There are 8 boys and 12 girls in a class. 5 students have to be chosen for an educational trip. Find the number of ways in which this can be done if 2 particular girls are always included

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 17

Answer – b) 816 Explanation : 18c3 = 816 (2 girls already selected)

Permutations And Combinations - MCQ 1 - Question 18

 In how many different ways the letters of the world INSIDE be arranged in such a way that all vowels always come together

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 18

Answer – b) 72 Explanation : Three vowels I, I and E can be arranged in 3!/2! Ways, remaining letters and group of vowels can be arranged in 4! Ways. So 4!*3!/2!

Permutations And Combinations - MCQ 1 - Question 19

How many 3 digit number can be formed by 0, 2, 5, 3, 7 which is divisible by 5 and none of the digit is repeated.

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 19

Answer – a) 24 Explanation : Let three digits be abc, a can be filled in 4 ways (2,3, 5 and 7) c can be filled in 2 ways (0 or 5) and b can be filled in 3 ways. So, 4*3*2 = 24 ways

Permutations And Combinations - MCQ 1 - Question 20

In a group of 6 boys and 8 girls, 5 students have to be selected. In how many ways it can be done so that at least 2 boys are included

Detailed Solution for Permutations And Combinations - MCQ 1 - Question 20

Answer – b) 1526 Explanation : 6c2*5c3 + 6c3*5c2 + 6c4*5c1 + 6c5

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