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Install Now**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. **Total number of digits = 9 and Number of digits used without repetition = 3

Number of permutations = = = 504

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

**Ans. **Total number of digits = 10 and Number of digits used without repetition = 4

0 cannot be filled in the fourth place therefore number of permutations for fourth place = 9

Number of permutations = = = 504

Hence total number of permutations = = 4536

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

**Ans. **Total number of digits = 6 and unit place can be filled with any one of the digits 2, 4, 6.

Number of permutations = = = 3

Now, the tens and hundreds places can be filled by remaining 5 digits.

Number of permutations = = = 20

Therefore, total number of permutations = = 60

**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?**

**Ans. **Total number of digits = 5 and Number of digits used without repetition = 4

Number of permutations = = = 120

Now, unit place can be filled with any one of the digits 2, 4 for even number.

Number of permutations = = = 2

Now, the remaining three places can be filled with the remaining 4 digits.

Number of permutations = = = 24

Therefore, total number of permutations of even numbers = = 48

**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?**

**Ans. **Total number of persons = 8 and Number of persons used without repetition = 2

Number of permutations = = = 56

**6. Find if **

**Ans. **Given:

**7. Find if: **

**(i) **

**(ii) **

**Ans. (i)** Given:

or

is not possible because , therefore,

**(ii)** Given:

or

is not possible because , therefore,

**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. **Total number of letters in word EQUATION = 8

Number of letters used (all different) = 8

Number of permutations = = = 40320

**9. How many words, with or without meaning can be formed using all 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 by first letter is a vowel?**

**Ans. **Total number of letters in word MONDAY = 6 and number of vowels = 2

**(i)** Number of letters used = 4

Number of permutations = = = 360

**(ii)** Number of letters used = 6

Number of permutations = = = 720

**(iii)** First letter is vowel.

Number of permutations of vowels = = = 2

Now, remaining five places can be filled with remaining five letters.

Number of permutations = = = 120

Therefore, total number of permutations = = 240

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

**Ans. **Total letters of the word MISSISSIPPI = 11

Here M = 1, I = 4, S = 4 and P =2

Number of permutations = = 34650

When the four I’s come together then it becomes one letter so total number of letters in the word when all I’s come together = 8

Number of permutations = = 840

Number of permutations when four I’s do not come together = 34650 – 840 = 33810

**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?**

**Ans. **Total numbers in the word PERMUTATIONS = 12 and here T = 2

**(i)** First letter is P and last letter is S which are fixed.

Therefore, remaining 10 letters are to be arranged between P and S.

Number of permutations = = 1814400

**(ii)** There are 5 vowels in the word PERMUTATIONS. All vowels can be put together.

Number of permutations of all vowels together = = = 120

Now consider the 5 vowels together as one letter. SO the number of letters in the word when all vowels are together = 8

Number of permutations = = 20160

**(iii)** P and S are on 1^{st} and 6^{th} places P and S are on 2^{nd} and 7^{th} places

P and S are on 3^{rd} and 8^{th} places P and S are on 4^{th} and 9^{th} places

P and S are on 5^{th} and 10^{th} places P and S are on 6^{th} and 11^{th} places

P and S are on 7^{th} and 12^{th} places

Now, P and S can be put in 7 ways and also P and S can interchange their positions.

Number of permutations = = 14

Now the remaining 10 places can be filled with remaining 10 letters.

Number of permutations = = 1814400

Therefore, total number of permutations = = 25401600

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