Question 1

Repetition of the digits is allowed?

Accepted Answer

There are as many ways of filling 3 empty places in succession by the given 5 digits. In the given case, repetition of digits is allowed. Thus, the units place can be filled in by any of the all 5 digits. Similarly, tens and hundreds digits can be filled by any of the all 5 digits. Thus, the number of ways in which 3-digit numbers can be formed from the given digits is 5×5×5=125.

Question 2

Repetition of the digits is not allowed?

Accepted Answer

In the given case, repetition of digits is not allowed. Thus, if units place is filled at first, then it can be filled by any of the given 5 digits. Thus, the number of ways is 5 . Similarly, the tens place can be filled by remaining 4 digits and the hundreds place can be filled by remaining three digits. Therefore, by multiplication principle, the number of ways = 5×4×3=60.

Question 3

How many 3-digit even numbers cab be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

Accepted Answer

Ans: There are ways of filling 3 empty places in succession by the given 6 digits. In the given case, the units place can be filled in by any of the all 6 digits. Similarly, tens and hundreds digits can be filled by any of the all 6 digits, as repetition is allowed. Thus, by multiplication principle, the required 3-digit even numbers is 3×6×6=108

Question 4

How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Accepted Answer

Ans: There are ways of filling 4 empty places in succession by the given 10 letters of the English alphabet, given that repetition is not allowed. The first place can be filled in 10 different ways any of the given 10 letters and second place can be filled in 9 different ways. Similarly third place can be filled in 8 different ways and the fourth in 7 different ways. Thus, by multiplication principle, the number of ways in which 4 empty places can be filled is 10×9×8×7=5040 Therefore, 5040 4-letter codes can be formed with the help of first 10 letters of English alphabet without repetition.

Question 5

How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

Accepted Answer

Ans: Given: 5-digit telephone number starts with 67. Therefore, there are ways of filling the 3 empty spaces 6, 7, __, __, __ by digits 0-9 without repetition. Thus, units place can be filled in 8 different ways, tens place can be filled in 7 different ways and hundreds place can be filled in 6 different ways. Thus, by multiplication principle, number of ways in which 5-digit telephone number can be constructed is 8×7×6=336

Question 6

A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Accepted Answer

Ans: When a coin is tossed once, there are 2 outcomes (head and tail) i.e., the number of ways of showing different face = 2 Thus, by multiplication principle, number of possible outcomes = 2×2×2=8

Question 7

Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Accepted Answer

Ans: 1 signal = 2 flags. There are ways of filling 2 empty places in succession by the given 5 flags. The upper empty place can be filled in 5 different ways and the lower empty place can be filled in 4 different ways. Thus, by the multiplication principle, the ways in which different flags can be generated =5×4=20

Question 8

8!

Accepted Answer

8! = 1×2×3×4×5×6×7×8=40320

Question 9

4! – 3!

Accepted Answer

4! = 1×2×3×4 = 24 3!=1×2×3 = 6 ∴4! - 3! = 24 - 6 = 18

Question 10

Is 4! + 3! = 7!?

Accepted Answer

3!=1×2×3 = 6 4! = 1×2×3×4 = 24 ∴3!+4!=6+24=30 7!=1×2×3×4×5×6×7=5040 ∴3!+4!≠7!

Question 11

How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Accepted Answer

Ans: The number of different letters in the given word is 8. Thus, the number of words than can be formed without repetition is number of permutations of 8 different objects taken 8 at a time = 8P8=8! Therefore, number of words formed = 8! = 40320.

Question 12

Words start with P and end with S.

Accepted Answer

Here, P and S are fixed at extreme ends (P at left and S at right end). Hence, 10 letters are left. Therefore, required number of permutations 10!/2!=1814400

Question 13

Vowels are all together.

Accepted Answer

Since, all the 5 vowels appear only once and they should always occur together, they can be treated as a single object. This single object with the remaining 7 objects will be 8 objects. Thus, 8 objects in which 2 T’s can be arranged together is 8!/2!ways Also, the five different vowels can be arranged in 5! ways. Thus, by multiplication principle, the number of arrangements =8!/2!×5!=2419200

Question 14

There are always 4 letters between P and S?

Accepted Answer

The letters can be arranged in such a way that there are 4 letters between P and S. Thus, letters P and S are fixed. The remaining 10 letters in which 2 T’s is arranged in 10!2!ways. Also, there are 4 letters between P and S = 2×7=14ways Thus, by multiplication principle, the number of arrangements = 10!/2!×14=25401600

Chapter 7 permutations and combinations Questions and Answers: NCERT Solutions for Class 11th Maths

Class 11 Maths Chapter 7: permutations and combinations - Questions and Answers of NCERT Book Solutions

Exercise 7.1

1. How many 3 -digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that (i) Repetition of the digits is allowed? (ii) Repetition of the digits is not allowed?

2. How many 3-digit even numbers cab be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

Ans: When a coin is tossed once, there are 2 outcomes (head and tail) i.e., the number of ways of showing different face = 2 Thus, by multiplication principle, number of possible outcomes = 2×2×2=8

6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

Exercise 7.2

1. Evaluate (i) 8!

Ans: (i)8! = 1×2×3×4×5×6×7×8=40320

(ii) 4! – 3!

Ans: 4! = 1×2×3×4 = 24 3!=1×2×3 = 6 ∴4! - 3! = 24 - 6 = 18

2. Is 4! + 3! = 7!?

Ans: 3!=1×2×3 = 6 4! = 1×2×3×4 = 24 ∴3!+4!=6+24=30 7!=1×2×3×4×5×6×7=5040 ∴3!+4!≠7!

3. Compute

5.Evaluate

(i) n = 6, r = 2

Ans: (i)Whenn = 6,r = 2:

(ii) n = 9, r = 5

(ii)Whenn = 9,r = 5:

Exercise 7.3

1. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digits is repeated?

Ans: Here, what matters is the order of digits. Thus, there are permutations of 9 different digits taken 3 at a time. Therefore, the number of 3-digit numbers =

2. How many 4-digit numbers are there with no digit repeated?

3. How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?

4. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?

5. From a committee of 8 persons, how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position.

Ans: In a committee of 8 persons, a chairman and vice chairman are selected in such away that one person can hold only one position. Thus, ways of choosing a chairman and vice chairman is permutation of 8 objects taken 2 at a time Therefore, number of ways

6. Find n if

7. Find r if

8. How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

Ans: The number of different letters in the given word is 8. Thus, the number of words than can be formed without repetition is number of permutations of 8 different objects taken 8 at a time = 8P8=8! Therefore, number of words formed = 8! = 40320.

9. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated if, (i) 4 letters are used at a time (ii) all letters are used at a time (iii) all letters are used but first letter is a vowel.

10. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

11. In how many ways can the letters of the word PERMUTATIONS can be arranged if the (i) Words start with P and end with S. (ii) Vowels are all together. (iii) There are always 4 letters between P and S?

Exercise 7.4

1.If

Ans:Since, we know that nCa=nCb⇒a=b or m=a+b That is,

2. Determine n if

3. How many chords can be drawn through 21 points on a circle?

Ans: To draw one chord, 2 points are needed. The number of combinations should be counted to find the number of chords to be drawn through the 21 points on the circle. Thus, there are combinations of 21 points taken 2 at a time. Therefore, number of chords is

4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls.

Ans: Given: 3 boys and 3 girls are to be selected from 5 boys and 4 girls. Out of 5 boys, 3 boys can be selected in 5C3 ways. Out of 4 girls, 3 girls can be selected in 4C3 ways. Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected =

5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls each colour.

6.Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Ans: There are 4 aces in a deck of 52 cards. A combination of 5 cards with one ace has to be made. One ace is selected in 4C3 ways and the remaining 4 cards is selected from the 48 cards in 48C4 ways. Therefore, by multiplication principle, Number of combinations is

7. In how many ways can one select a cricket team of 11 from 17 players in which 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Ans: There are 5 black and 6 red balls in the bag.2 black balls can be selected out of 5 black balls in 5C2ways and 3 red balls can be selected out of 6 red balls in 6C3ways. Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls

9.In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Ans: Given: Out of 9 courses, 2 courses are compulsory. Thus, out of the remaining 7 courses, every student has to choose 3 courses. This can be chosen in 7C3 ways.Thus, required number of ways of choosing the programme is:

1. How many 3 -digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(i) Repetition of the digits is allowed?
(ii) Repetition of the digits is not allowed?

9. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated if,
(i) 4 letters are used at a time
(ii) all letters are used at a time
(iii) all letters are used but first letter is a vowel.

11. In how many ways can the letters of the word PERMUTATIONS can be arranged if the
(i) Words start with P and end with S.
(ii) Vowels are all together.
(iii) There are always 4 letters between P and S?