**Q-1: If there is no repetition of numbers, then find the total of 3- digit numbers formed by using digits from 2 to 9.**

** **

**Solution:**

**Digits** in **between** **2 to 9 are 8.**

We have to construct 3- digit numbers by using the digits from 2 to 9.

It is given that; the **repetition of digit is restricted**.

So, from the **permutation** point, there are **permutations** of 8 digits ( different ) among which we need to take 3 at a time.

Hence, **the total of 3- digit number formed by using digits from 2 to 9** =

= **= 336**

**Therefore, total of 336 3-digit numbers can be formed by using digits from 2 to 9 without any repetition.**

**Q-2: Find the total number of 4- digits numbers where there is no repetition of digits.**

** **

**Solution:**

We have to construct 4- digit numbers from 0 to 9.

**Repetition of digit is restricted.**

We know that to have a proper number the thousands place must be filled with **9 digits from 1 to 9**, because **0** will be considered as **negligible** at **thousands place**.

Therefore, the **total number of ways** through which the **thousands** **place** should be filled **is 9.**

Now, the hundreds place must be filled with remaining **9 digits between 0 to 9**, except that number which is inserted at thousands place, as the **repetition is restricted**. Similarly, **tens and ones place** can be filled by the **remaining 9 digits.**

Hence, for hundreds, tens and ones, there will be 3- digit number and therefore, the **permutations** of 9 different digits will be taken **3 at the same time**.

Now,

**The total number of 3- digit number:**

=

=

**Therefore, by the principle of the multiplication, the total number of the 4 – digit number is 504 × 9 = 4536**

**Q-3: Find the total number of 3- digit number which is even and is formed by using the digits 2, 3, 4, 5, 6, 7 and 8. Note that repetition of digit is restricted.**

** **

**Solution:**

We need to form even numbers of **3- digits** which can be formed by using the digits such as **2, 3, 4, 5, 6, 7 and 8.** **Repetition of the digit is restricted.**

To make the **3- digit number** **even**, the **unit place** must be filled with an **even number** that is **2, 4, 6, and 8** among the given number.

As it is given in the question that the **repetition of the numbers** is restricted so, the **unit** **place** must be already occupied with an even number, and the **hundreds** and **tens** **place** is to be filled with the **remaining 6 digits** given in the question.

Hence, to fill up the **hundred** and **tens place** we need to pick up digits except that which is being filled on **ones place**. So, we need to pick with **permutation** **of 6 different digits** among which **2 digits** are **taken** at a time.

Hence, the **number of ways through which the hundreds and tens place is filled:**

=

= **= 30**

**Therefore, by the principle of the multiplication, the number of the 3- digit numbers required is 4 × 30 = 120.**

**Q-4: How many 4- digit numbers will be formed by using the digits 2, 3, 4, 5 and 6. Note that, repetition of digits is not allowed. Find among those numbers, how many numbers will be even number?**

** **

**Solution:**

**We need to construct 4 – digit numbers using the digits 2, 3, 4, 5 and 6.**

We have **5 digits** among which we need to form **4- digit** number as there are **permutations of 5 different digits taken 4 at a time.**

Hence, the required number of the 4- digit numbers:

=

= **= 120**

Among these **120 numbers** of **4 – digit** which is formed by using **2, 3, 4, 5 and 6**, even numbers must end with an even number which is **2, 4, or 6**.

There are chances that there are **3! Number** of ways through which the ones place is filled with **digits**, i.e., it can be filled by **6 number of ways.**

As, in the question, it is mentioned that repetition of **digits are restricted** and the **units** **place** is already occupied with an **even** **number**, so the remaining place is to be filled by the **remaining 4 digits** which is not inserted at any place.

Hence, the number of ways by which the **thousands**, **hundreds** and **tens** **place** will be filled is the **permutation** of **4 different digits** which is taken **3 at a time**.

**Therefore, the number of ways to fill the remaining places**:

=

= **= 24**

**Hence, by the principle of multiplication, the required number which is an even number is 24 × 3 = 72.**

**Q-5: Consider a committee with 9 persons, in how many ways can we choose a chairman and a vice chairman? Note that, a person can hold only one position at a time.**

** **

**Solution:**

There is a **committee** with **9 persons** among which, **a vice- chairman and a chairman** is chosen in such a way that **one position can be held by only one person**.

So,

To choose a **chairman** and **a vice- chairman**, there are number of ways with **permutation** of **8 different persons** **taken 2 at a time.**

**Therefore, the number of ways required:**

=

= **= 72**

**Q-6: Find n, if **

** **

**Solution:**

**Therefore, n = 9**

**Q-7: Find the value of a, if:**

**(i). 5Pa=2[6Pa–1] (ii). 5Pa=6Pa–1**

** **

**Solution:**

**(i).**

**So, ( 7 – a ) ( 6 – a ) = 12 **

**So, a = 3 or a = 10**

**It is known that,**

**Therefore, a ≠ 10**

**Hence, r = 3**

**(ii).**

**So, ( 7 – a ) ( 6 – a ) = 6**

**So, a = 4 or a = 9**

**It is known that,**

**Therefore, a ≠ 9**

**Hence, a = 4**

**Q-8: Get all the number of ways, with or without the meaning, which might be formed by using the letters from the word EQUATION. Note that, repetition of the word is restricted.**

** **

**Solution:**

In the word **EQUATION**, we have **8** **letters** from the **English** **alphabets**.

Hence, to find the number of words formed by using all those 8 letters of the word **EQUATION**, without repetition of the alphabets, we have:

**Permutation** of **8 different letters taken 8 at a time**, which is **= 8!**

**Hence, the number of words required which can be formed = 8! = 40320.**

** **

**Q-9: Find the number of words, with or without the meaning, which might be formed by using the letters from the word MONDAY. Note that, repetition of the word is restricted, if**

**(i) At a time, 3 letters can be used.**

**(ii) All the letters can be used at a time.**

**(iii) First letter is a vowel and all the letters are being used.**

** **

**Solution:**

In the word **MONDAY**, we have **6 letters** from the **English** **alphabets**.

**(i)** As per the question demand, we need to use 3 letters at a time.

So, the number of words formed by the letters of the word MONDAY, without the repetition of the letters, is

The permutation of 6 different objects which are taken 3 at a time is given by,

Hence, the number of words required which can be formed by using the word MONDAY using only 3 letters at a time is given by,

**120 words**

**(ii)** In this case, at a time, we can use all the letters of the given word **MONDAY**.

So, to find the number of words formed, we have:

**Permutation** of 6 different letters taken any of the **6 at a time**, that is:

Hence, **the number of words required** which can be formed by using all the letters of the word **MONDAY** at a time = **6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.**

**(iii)** In the given words **MONDAY**, we have two vowels which is **O and A.**

As per the given condition, vowels must be filled at the start of the word. Also, the **repetition of letter is restricted.**

So, to this can be done only in **2! Number of ways**.

We have **6 letters** and the **repetition is restricted**, also the rightmost place will be filled by a **vowel** from the **word** **MONDAY**, so the remaining **5 places** will be filled by the remaining **5 letters**. And, this can be done in **5!** **Number of ways.**

**Hence, in this case, the number of words required and being formed is 5! × 2! = 5 × 4 × 3 × 2 × 1 × 2 × 1 = 240.**

**Q-10: Find the number of distinct permutations such that 4 I’s does not come together in the word MISSISSIPPI.**

**Solution:**

In the word **MISSISSIPPI, M appears for 1 time, I appears for 4 times, P appears for 4 times and S appears for 4 times.**

So, the required number of the distinct **permutations** of the letters in the word **MISSISSIPPI** is given by:

**= 34650.**

As there are **4 I’s** in this given number which is **MISSISSIPPI**, whenever they will occur together, they will be treated as a single object which is **IIII.** Since, all the **4 I’s** are considered as **1** and there are **7 letters** left in the given word **MISSISSIPPI**, so we have total of **8 words** in account.

Among these **8 words** **S **appears for **4** **times** and **P** appears for **4 times**, and **M** appears just **once**.

So, this can be arranged in **840 ways**

So, the **number** of **arrangements** where **all I’s** can occur together **= 840**

**Hence, the total number of the distinct permutations of the letters in the given word MISSISSIPPI, where all the 4 I’s won’t come together = 34650 – 840 = 33810.**

**Q-11: Arrange the letters of the word PERMUTATIONS and observe the number of ways of the permutation if,**

**(i). the new word starts with P and ends with S.**

**(ii). when the vowels are taken all together.**

**(iii). there will always be 4 letters between P and S.**

** **

**Solution:**

In the given word **PERMUTATIONS**, all the other letters occurred for once except **T** which occurred for **twice**.

**(i)** As we need to fix **P** and **S** at the extreme ends i.e., **P** at the **left** **side** of the **word** and **S** on the **right** **side** of the **word**, then we have **10** more letters to fix.

**Therefore, number of arrangements required in this case = 10!2! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 = 1814400**

** **

**(ii)** In the given word **PERMUTATIONS**, we have all the **5** **vowels**, i.e**., a, e, i, o, u** and each of the vowel is appearing only for once in the given word.

As in this case, they have to occur together, so these vowels are considered as a single object. Since, we have **12 letters** in the given word and as all the **vowels** are counted as one, so we have more **7** **letters**. So, **8 objects** are there in the account.

Now,

All these letters can be arranged in **ways** where there is **2 T’s**.

As there are 5 different vowels, then corresponding to these arrangements, **vowels** will be arranged in **5! Ways.**

**Hence, by the principle of multiplication, number of arrangements required for this case = 8!2!×5! = 2419200.**

**(iii)** We have to arrange the letters of the word **PERMUTATIONS** in such a way that always there will be **4 letters in between P and S.**

So, it is clear that the **position of P and S** is fixed and **4** others letters can be inserted between **P and S**. There are **2 T’s** in the remaining **10 letters** which can be arranged in

Also, **4** **letters** can be placed in between P and S in **2 × 7 = 14 ways.**

**Hence, by the principle of multiplication, number of arrangements required in this case = 10!2!×14 = 25401600.**