Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3

Question 1.

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

Solution:

3-digit numbers are to be formed.

i.e., Three places are to be filled with digits 1 to 9.

This can be done in 9 × 8 × 7

= 504 ways,

Question 2.

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

Solution:

Let the digits be 0 to 9.

4-digit numbers are ^{10}P_{4}.

This includes those numbers which have 0 in the beginning (one thousand’s place)

3-digit numbers out of 9 digits 1-9 are ^{9}P_{3}.

∴ 4-digit numbers which do not have 0 in the beginning (on the extreme left)

= ^{10}P_{4} – ^{9}P_{3}

= 10 × 9 × 8 × 7 – 9 × 8 × 7

= 9 × 8 × 7(10 – 1)

= 9 × 9 × 8 × 7

= 4536.

Question 3.

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

Solution:

Let 2 be fixed at unit’s place.

Now, we have 5 digits and 3 places are to be filled up.

This can be alone in ^{5}P_{2} ways.

When unit’s place is filled up both 4 or 6, again in each case, we have ^{5}P_{3} numbers.

∴ 4-digit even numbers are

3 × ^{5}P_{2} = 3 × 5 × 4

= 60.

Question 4.

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

Solution:

(i) Out of 5 digits, 4-digit numbers are to be formed.

Such numbers are = ^{5}P_{4} = 5 × 4 × 3 × 2

= 120.

(ii) When 2 is at unit’s place, then remaining three places are filled in ^{4}P_{3} ways

= 4 × 3 × 2

= 24.

When 4 is at unit’s place, then 4-digit numbers are again = 24.

∴ Even 4-digit numbers

= 2 × 24

= 48.

Question 5.

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

Solution:

Out of 8 persons, chairman can be chosen in 8 ways. After selecting a chairman, we have to choose a vice chairman out of 7 persons.

This can be done in 7 ways.

Out of 8 persons, a chairman and a vice chairman can be chosen

8 × 7 = 56 ways.

Question 6.

Find n, if ^{n-1}P_{3} : ^{n}P_{4} = 1 : 9.

Solution:

or n = 9.

Question 7.

Find r, if

(i) ^{5}P_{r} = 2^{6}P_{r-1}

(ii) ^{5}P_{r} = ^{6}P_{r-1}

Solution:

(i) ^{5}P_{r} = 2^{6}P_{r-1}

or (7 – r)(6 – r) = 12.

⇒ 42 – 13r + r^{2} = 12 or r^{2} – 13r + 30 = 0.

⇒ (r – 10)(r – 3) = 0 ⇒ r = 3 or 10.

r ≠ 10, since r cannot be greater than n.

∴ r = 3.

(ii) ^{5}P_{r} = ^{6}P_{r-1}

or (7 – r)(6 – r) = 6 or 42 – 13r + r^{2} = 6.

or r^{2} – 3r + 36 or (r – 9)(r – 4) = 0.

⇒ r = 9 or 4. So, r = 4, since r ≠ 9, because r cannot be greater than 4.

Question 8.

How many words can be formed using all letters of the word EQUATION, using each letter exactly once?

Solution:

Number of letters in equation = 8.

∴ Number of letters to be taken at a time = 8.

If p is the number of words thus formed, then

P = P(8, 8) = \(\frac{8!}{(8 – 8)!}\) = 8!

= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1

= 40320.

Question 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?

Solution:

(i) MONDAY has 6 letters. 4 letters are taken at a time.

∴ Number of words = ^{6}P_{4}

= 6 × 5 × 4 × 3

= 360.

(ii) All the letters of word MONDAY are taken at a time.

Number of words = 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

(iii) Let the words begin with A.

Number of words formed with 5 letters = 5! = 120

Similarly, when words begin with ‘O’, number of such words = 120.

∴ Number of words beginning with vowel

= 2 × 120

= 240.

Question 10.

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

Solution:

In the given word their are 4I, 4S, 2P and 1M.

Total number of permutations with no restriction = \(\frac{11!}{4!4!2!}\)

If we take 4I as one letter, then total letters become = 11 – 4 + 1 = 8.

If p is the permutations, when 41s are not together, then

= 34650 – 840

= 33810.

Question 11.

In how many ways, can the letters of the word PERMUTATIONS be arranged, if the

(i) Words start with P and end with S?

(ii) Vowels are together?

(iii) There are always 4 letters between P and S?

Solution:

(i) Letters between P and S are ERMUTATION. These 10 letters having T two times. These letters can be arranged in \(\frac{10!}{2!}\) ways = 1814400 ways.

(ii) There are 12 letters in the word PERMUTATIONS which have T two times.

Now the vowels a, e, i, o, u are together.

Let it be considered in one block.

The letters of vowels can be arranged in 5! ways.

Thus, there are 7 letters and 1 block of vowels with T two times.

∴ Number of arrangements = \(\frac{8!}{2!}\) × 5! = \(\frac{8!5!}{2}\) = 2419200.

(iii) There are 12 letters to be arranged in 12 places.

These 12 letters are filled in 12 places as shown above.

P may be filled up at place No. 1, 2, 3, 4, 5, 6, 7 and consequently S may be filled up at place No. 6, 7, 8, 9, 10, 11, 12 leaving four places in between. Now P and S may be filled up in 7 ways. Similarly, S and P may be filled of in 7 ways.

Thus, P and S or S and P can be filled up is 7 + 7 = 14 ways.

Now remaining 10 places can be filled by E, R, M, U, T, A, T, I, O, N in \(\frac{10!}{2!}\) ways.

∴ No. of ways in which 4 letters occur between P and S

= \(\frac{10!}{2!}\) × 14 = \(\frac{3628800}{2}\) × 14 = 2540100.