NCERT SOLUTIONS FOR 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.

Total digits are 9. We have to form 3 digit numbers without repetition.

∴ The required 3 digit numbers = 9P3

=\frac { 9! }{ 6! } =9\times 8\times 7=504

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

Solution.

The 4-digit numbers are formed from digits 0 to 9. In four digit numbers 0 is not taken at thousand’s place, so thousand’s place can be filled in 9 different ways. After filling thousand’s place, 9 digits are left. The remaining three places can be filled in 9P3 ways.

So the required 4-digit numbers

= 9 x 9P3

= 9 x 504 = 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.

For 3-digit even numbers unit place can be filled by 2, 4, 6 i.e in 3 ways. Then the remaining two places can be filled in 5P2 ways.

∴ The required 3-digit even numbers

= 3 x 5P2

= 60

Question 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?

Solution.

The 4-digit numbers can be formed from digits 1 to 5 in 5P4ways.

∴ The required 4 digit numbers = 5P4 = 120 For 4-digit even numbers unit place can be filled by 2,4, i.e., in 2 ways. Then the remaining three places can be filled in 4P3 ways.

∴ The required 4-digit even numbers

= 2 x 4P3 = 2 x 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.

From a committee of 8 persons, we can choose a chairman and a vice chairman

Question 6. Find n if  n-1P3nP4 = 1 : 9.

Solution.

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3 1

Question 7. Find r if

  • (i) 5Pr = 26Pr-1
  • (ii) 5Pr = 6Pr-1

Solution.

NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3 2

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

Solution.

No. of letters in the word EQUATION = 8

∴ No. of words that can be formed

8P8 = 8!

=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.

No. of letters in the word MONDAY = 6

(i) When 4 letters are used at a time.

Then, the required number of words

6P4

=\frac { 6! }{ 2! } =6\times 5\times 4\times 3=360

(ii) When all letters are used at a time. Then the required number of words
6P6 = 6!
= 720

(iii) All letters are used but first letter is a vowel.
So the first letter can be either A or O.
So there are 2 ways to fill the first letter & remaining places can be filled in 5P5 ways.
∴ The required number of words
= 2 x 5P5
= 2 x 5! =240.

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

Solution.

There are 11 letters, of which I appears 4 times, S appears 4 times, P appears 2 times & M appears 1 time.

∴ The required number of arrangements

=\frac { 11! }{ 4!4!2! } =\frac { 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4! }{ 4\times 3\times 2\times 2\times 4! }

= 10 x 10 x 9 x 7 x 5 = 34650                       … (i)

When four I’s come together, we treat them as a single object. This single object with 7 remaining objects will account for 8 objects. These 8 objects in which there are 4S’s & 2P’s

can be rearranged in \frac { 8! }{ 4!2! }  ways i.e. in 840 ways      … (ii)

Number of arrangements when four I’s do not come together = 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 all together,
  • (iii) there are always 4 letters between P and S?

Solution.

There are 12 letters of which T appears 2 times

(i) When words start with P and end with S, then there are 10 letters to be arranged of which T appears 2 times.

∴ The required words = \frac { 10! }{ 2! }

=\frac { 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2! }{ 2! } =1814400

(ii) When vowels are taken together i.e E U A I O we treat them as a single object. This single object with remaining 7 objects will account for 8 objects, in which there w are 2Ts, which can be rearranged in \frac { 8! }{ 2! } =20160 ways.

Corresponding to each of these arrangements the 5 vowels E, U, A, I, O can be rearranged in 5! = 120 ways. Therefore, by multiplication principle, the required number of arrangements = 20160 x 120 = 2419200.

(iii) When there are always 4 letters between P & S

∴ P & S can be at

  • 1st & 6th place
  • 2nd & 7th place
  • 3rd& 8th place
  • 4th & 9th place
  • 5th & 10th place
  • 6th & 11th place
  • 7th & 12th place.

So, P & S will be placed in 7 ways & can be arranged in 7 x 2! = 14

The remaining 10 letters with 2T’s, can be arranged in \frac { 10! }{ 2! } =1814400 ways.

∴ The required number of arrangements = 14 x 1814400= 25401600.

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