CBSE Class 11th Math 7 – Permutations and Combinations MCQs

The number of ways in which letters of the word PARENT can be arranged so that R is always next to A is ____.

Correct! Wrong!

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Let us keep AR together and consider it as one letter. Now, we have 5 letters which can be arranged in a row in 5P5 = 5! = 120 ways. The total number of ways in which R is always next to A is 120.

How many three digit even numbers can be formed by using the digits 3, 4, 5, 6, 7, 8 when i) repetition of digits is allowed? ii) Repetition of digits is not allowed?

Correct! Wrong!

i) As repetition of digits is allowed, the ten's place can be filled up by any of the 6 digits. So there are 6 ways of filling the ten's place. The hundred's place can also be filled by 6 ways as repetition of digits is allowed. For the number to be even, the unit's place can be filled by 4, 6, 8. There are three ways of filling the unit's place. Hence the required number of 3 digit even numbers = 3 × 6 × 6 = 108 ii) The unit's place can be filled in 3 ways (by 4, 6, 8.) The ten's place can be filled by any of the remaining 5 digits in 5 ways as repetition of digits is not allowed. The hundred's place can be filled by the remaining 4 digits in 4ways. Hence the required number of 3 digit numbers =3 × 5 × 4 = 60

There are 5 items in column I and 5 in column II. A student is asked to match each item in column I with an item in column II. The number of possible (correct and incorrect both included) answers are there in this question is ____.

Correct! Wrong!

In column I, the first item can be matched with any of the 5 items in column II. This can be done in 5 ways. When this has been done, the second item in column I can be matched with any of the remaining 4 items of column II.- The number of ways of matching the third, fourth and the fifth of column I are 3, 2, 1 respectively. Hence the required number = 5 × 4 × 3 × 2 × 1 =120

The number of different products can be obtained by multiplying 2 or more of the numbers 2, 3, 6, 7 without repetition is ____.

Correct! Wrong!

Number of products = Number of ways of selecting 2 or 3 or all out of 4 numbers 2, 3, 6, 7. = 4C2 + 4C3 + 4C4 = 6 + 4 + 1 = 11

A committee of 4 is to be formed out of 5 men and 3 ladies. In how many ways can this be done, when atleast 2 ladies are included?

Correct! Wrong!

We have to make a selection of (i) 2 ladies out of 3 and 2 men out of 5; or (ii) 3 ladies out of 3 and 1 man out of 5 The number of ways of these selections are: (i) 3C2 × 5C2 = 30 (ii) 3C3 × 5C1= 5 ∴ the required number of ways = 30 + 5 = 35

There are 7 items in column I and 7 items in column II. A student is asked to match each item in column I with an item in column II. The number of possible, correct or incorrect answers is ____.

Correct! Wrong!

Each answer to the given question is an arrangement of the 7 items of column II, keeping the order of items in column I fixed. The total number of answers = number of arrangements of 7 items in column II. = 7P7 =7!

The number of different numbers greater than 60000 that can be formed with the digits 0, 2, 2, 6, 8 is ____.

Correct! Wrong!

Numbers greater than 60000 will have either 6 or 8 in the first place and will consist of 5 digits. Number of numbers with digit 6 at first place = 4!/2! Number of numbers with digit 8 at first place = 4!/2! ∴ The required number of numbers = 4!/2! + 4!/2! = 24

A girl can take 3 books from her school library and there are 10 books of her interest in the library. Of these 10, she does not want to borrow Mechanics Part II, unless she borrows Mechanics Part I. In how many ways can she choose the 3 books to be borrowed?

Correct! Wrong!

These are the possibilities: (i) When Mechanics Part I is borrowed, In this case the girl may borrow Mechanics Part II. So, she has to choose 1 book out of the remaining 8 books of her interest. This can be done in 8C1 ways. (ii) When Mechanics Part I is not borrowed, In this case the girl does not want to borrow Mechanics Part II. So, she has to choose 3 books from the remaining 8 books. This can be done in 8C3 ways. Hence the required number of ways = 8C1 + 8C3= 64

The number of ways in which 7 cups can be arranged on 5 saucers is ____.

Correct! Wrong!

The number of ways in which 7 cups be arranged on 5 saucers is the same as the number of arrangements of 7 things taking 5 at a time. The required number = 7P5 = 7!/7-5! = 7!/2! = 2520

With the digits 0, 1, 2, 4, 5, 6 and with no digit being repeated, the number of numbers that can be formed between 300 and 900 is ____.

Correct! Wrong!

Numbers between 300 and 900 consist of 3 digits with digit at hundred's place greater than or equal to 3. Hundred's place can be filled by using the digits 4, 5, 6 in 3 ways. The ten's and unit's places can be filled by the remaining 5 digits in 5P2 ways. The required number of numbers = 3 × 5P2 = 3 ×5!/3! = 3 × 20 = 60

In how many ways can 4 pens be distributed among 5 girls, when i) no girl gets more than one pen? ii) a girl may get any number of pens? iii)no girl gets all the pens?

Correct! Wrong!

i) The first pen can be given in 5 ways as it may be given to any one of the 5-girls. The second pen can be given in 4-ways as the girl who got the first pen cannot receive the second pen. The third pen can be given in 3 ways and the fourth one in 2 ways. So, the number of ways in which all the pens can be given away= 5 × 4 × 3 × 2 =120 ii) The first pen can be given away in 5-ways as it may be given to anyone of the 5 girls. The second pen can also be given in 5 ways as it may be obtained by the girl who has already received a pen. Similarly, third and fourth pens can be given in 5 ways each. Hence the number of ways in which all the pens can be given away = 5 × 5 × 5 × 5 = 625 iii) As any one of the 5 girls may get all the pens, the number of ways in which a girl gets all the 4 pens 5 So the number of ways in which a girl does not get all the pens = 625 – 5 = 620

A bag contains 2 white marbles, 3 black marbles and 4 red marbles. In how many ways can 3 marbles be drawn from the box if atleast one black marble is to be included in the draw?

Correct! Wrong!

The 4-digit numbers can be formed without using the digits 2, 3, 0, 7, 9 is ____.

Correct! Wrong!

We have to form 4-digit numbers by using the digits 1, 4, 5, 6, 8.Clearly, the repetition of digits is allowed. So, each one of the unit’s place, ten's place and hundred's place can be filled by any of the digits 1,4,5,6,8. Each place can be filled in 5 ways. Total number of required numbers = 5 × 5 × 5 × 5 = 625

When a group photograph is taken, all the seven vice-presidents should be in the first row and all the thirty managers should be in the second row. The two corners of the second row are reserved for the two senior most managers and are interchangeable only between them. The middle seat of the front row is reserved for the shortest vice- President. The number of possible seating arrangements is ____.

Correct! Wrong!

As the middle seat in the front row is reserved for the shortest vice-president, the remaining 6 vice presidents can be arranged in 6P6 ways = 6! Ways. The two corners of the second row are reserved for the senior most managers who can occupy these 2 places in 2! Ways. The remaining 28 seats may be occupied by the remaining 28 managers in 28! Ways. By fundamental principle of counting, the total number of arrangements = 6!?(28!?2!)

The number of ways in which 7 men and 7 women can be seated in a row so that no two women may sit together is ____.

Correct! Wrong!

7 men can be seated in a row in 7P7 = 7! ways. Now, after arranging 7 men. 7 women can be arranged in 8P7 ways in the 8 gaps. The number of ways in which no two women sit together = 7! × 8P7= 7! × 8!

How many numbers are there between 100 and 1000 such that every digit is either 5 or 7?

Correct! Wrong!

We have to form 3-digit numbers by using 5 and 7. The repetition of digits is allowed. Each one of the unit's, ten's and hundred's places can be filled in 2 ways. Hence required number of numbers = 2 × 2 × 2 = 8

How many numbers are there between 100 and 1000 such that atleast one of their digits is 6?

Correct! Wrong!

Total number of 3-digit numbers having atleast one of their digits as 6 = (Total number of 3 digit numbers) − (Total number of 3-digit numbers in which 6 does not appear at all). Total number of 3 digit numbers = 9 × 10 × 10 = 900 (as the numbers have to be formed with the digits 0,1,2,......9 and∵ 0 cannot be used in the hundreds digit) Total number of 3 digit numbers in which 6 does not appear at all = 8 × 9 × 9 ( as the numbers have to be formed with the digits 0,1,2...9 except 6) [Each of the three digits will have one choice lesser than in the previous step, as 6 has to be excluded from each of the possibilities] Hence the required number of numbers =( 9 × 10 × 10) – (8 × 9 × 9) = 252

The number of numbers that can be formed with the digits 7, 6, 9, 8, 9, 6, 7 so that the odd digits always occupy the odd places is ____.

Correct! Wrong!

There are four odd digits 7, 9, 9, 7 and four odd places. These odd digits can be arranged in odd places in 4!/2!?2! ways. The 3 even digits 6, 8, 6 can be arranged in 3 even places in 3!/2! ways. ∴ the required number of numbers = 4!/2!2! × 3!/2! = 18

How many numbers are there between 100 and 1000 which have exactly one of their digits as 4?

Correct! Wrong!

a) 3-digit numbers with 4 at the hundred's place but neither at the unit's place nor at the ten's place: Number of ways to fill the hundred's place=1(by 4 only) Number of ways to fill the ten's place=9(by the digits from 0 to 9 except 4) Number of ways to fill the unit's place=9(by the digits from 0 to 9 other than 4) Required number=1 × 9 ×9=81 b)3-digit numbers with 4 at the ten's place but neither at the hundred’s place nor at the unit's place: Number of ways to fill the ten's place=1 Number of ways to fill the unit's place=9(by any of the digits from 0 to 9 except 4) Number of ways to fill the hundred's place=8(by any of the digits from 1 to 9 except 4) Required number = 1 × 9 × 8=72 c) 3-digit numbers with 4 at the unit's place but neither at the ten's place nor at the hundred's place. Number of ways to fill the unit's place(by 4 only)=1 Number of ways to fill the ten's place=9 (by any of the digits 0,1,2,3,5,6,7,8,9) Number of ways to fill the hundred's place=8 (by any of the digits 1,2,3,5,6,7,8,9) Hence the required number=1 × 9 × 8 =72 The total number of required type of numbers = 81 + 72 + 72 = 225

There are five routes between Chennai and Mysore. In how many different ways can a person go from Chennai to Mysore and return, i) if for returning any of the routes is taken? ii) if for returning the same route is taken ? iii) the same route is not taken for returning ?

Correct! Wrong!

i) He may take any route for going from Chennai to Mysore. So, there are five ways of going from Chennai to Mysore. After that, he may return by any of the routes --in five different ways. Hence total number of ways of going to Mysore and returning back to Chennai = 5 × 5 =25. ii) There are five ways of going to Mysore and there is only one way of returning --by the same route. Hence total number of ways = 5 × 1 = 5. iii) There are 5 ways of going to Mysore. But the person may not return by the same route. There are four ways of returning back to Chennai. Hence the required number of ways = 5 × 4 = 20.

The total number of arrangements of the letters in the expression p4q2r3 when written in full length is ____.

Correct! Wrong!

There are 4 p's, 2 q's and 3 r's. ∴ The total number of arrangements = 9!/4!2!3! = 1260

From 4 boys and 8 girls, the number of ways can 6 be chosen, (i) to include only one boy. (ii) to include atleast one boy ?

Correct! Wrong!

(i) Number of ways of choosing 6 to include only 1 boy = 4C1 × 8C5 = 224 (ii) Number of ways of choosing 6 to include atleast 1 boy = Total number of ways – Number of ways of selecting no boy. = 12C68C6 = 924 – 28 = 896

How many arithmetic progressions with 10 terms are there whose first term is in the set {1, 5, 7} and whose common difference is in the set {5, 7, 9}?

Correct! Wrong!

There are 3 ways to choose the first term and corresponding to each such way there are 3 ways of choosing the common difference. Hence the required number of arithmetic progressions is 3 × 3 = 9

The number of ways can 18 identical spoons and 16 identical forks be placed in a row on a table so that 2 forks may not be together is ____.

Correct! Wrong!

As no two forks should be together, we must first arrange all the spoons in only one way as shown: × S × S × S × ......× S × S × S denotes spoon and × denotes fork. Since there are 18 spoons, the number of places marked × are 19. Now, 16 forks are to be arranged in these 19 places so that no two forks are together. This can be done in 19C16ways. Since all the spoons are identical, they can be arranged in only 1 way. Hence the required number of ways = 1 × 19C16 × 1 = 969

The number of ways can 7 gifts be given to 8 boys studying in the same class is ____.

Correct! Wrong!

There are 7 gifts which have to be distributed among 8 boys. Each of these gifts can be given in 8 ways. So the required number is 8 × 8 × 8 × 8 × 8 × 8× 8 = 87 ways.

CBSE Class 11th Math 7 - Permutations and Combinations MCQs
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