**Q-1: Find the number of words, having meaning or meaningless, each of 2 vowels and 4 consonants will be formed by the letters of the word DAUGHTER?**

** **

**Solution:**

In **DAUGHTER** word, we have **3 vowels** which is **A, U and E**, rest **5 letters** are **consonants** which are **D, G, H, T and R.**

Now,

The number of way to select **2 vowels** at a time from **3 vowels** will have **permutation** say, ^{3}C_{2 }= 3

Also, the number of way to select **4 consonants** at a time from **5 consonants** will have **permutation** say, ** ^{5}C_{4}**= 5

Hence, the total number of combinations of **2 vowels** and **4 consonants** = 3 × 5 **= 15**

These **15 combinations** of **2 vowels** and **4 consonants** will be arranged in **6! number of ways.**

**Therefore, the number of different words required = 15 × 5! = 1800**

**Q-2: Find the number of words, which can be either meaningful or meaningless, which will be formed by using all of the letters of the word EQUATION at a time such that both the vowel and the consonant will occur together.**

** **

**Solution:**

In the given word **EQUATION**, we have **5 vowels**, that are **E, U, A, I and O,** and there are **3 consonants**, that are **Q, T and N.**

As all the **vowels** and **consonants** will occur together, means both **(A, E, I, O, U) and (Q, T, N)** will be considered as a **single letter**.

Thus, the **permutation** of **vowels and consonants** taken all at a single time is counted which is: ^{2}P_{2 }= 2!

Now, at each of the **permutation **for vowels, there is a **permutation of 5!** Taken all at a time and for consonants, the **permutation is 3!** taken all at a time.

**Therefore, by the principle of multiplication, number of the words required = 2! × 5! × 3! = 1440**

**Q-3: There are 10 boys and 5 girls through which a committee with 7 members has to be formed. Now, find the number of ways in which this might be created when the committee formed consists of:**

**(i) Exactly 4 girls (ii) at least 4 girls (iii) almost 4 girls?**

** **

**Solution:**

**(i)** **From 10 boys and 5 girls, a committee of 7 members is being formed.**

As in this case, exactly **4 girls** is needed to be there in the committee, so each committee should contain ( 7 – 4 ) = **3 boys** in the committee.

**Therefore, number of ways required in this case:**

= ^{5}C_{4 }× ^{10}C_{3}

=

**= 600 number of ways**

** **

**(ii)** Here, in this case, at least **4 girls** can be there in each committee.

So, in such cases, a committee can consist of

**(a) Either 3 boys or 4 girls (b) either 2 boys or 5 girls.**

**(a) 4 boys and 3 girls will be selected in ^{10}C_{3 }× ^{5}C_{4 }number of ways.**

**(b) 2 boys and 5 girls will be selected in ^{10}C_{2 }× ^{5}C_{5 }number of ways.**

**Hence, the number of ways required in this case:**

=^{ 10}C_{3 }× ^{5}C_{4 }+ ^{10}C_{2 }× ^{5}C_{5}

=

=

**= 645 number of ways.**

**(iii)** Here, in this case, almost **4 girls** can be there in each of the committee.

So, in this case, a committee can consist of:

**(a) 4 girls and 3 boys (b) 3 girls and 4 boys**

**(c) 2 girls and 5 boys (d) 1 girl and 6 boys**

**(e) No girl and 7 boys**

** **

**(a) 4 girls and 3 boys** will be selected in ** ^{5}C_{4 }× ^{10}C_{3 }**number of ways.

**(b) 3 girls and 4 boys** will be selected in ** ^{5}C_{3 }× ^{10}C_{4 }**number of ways.

**(c) 2 girls and 5 boys** will be selected in ** ^{5}C_{2 }× ^{10}C_{5 }**number of ways.

**(d) 1 girl and 6 boys** will be selected in ** ^{5}C_{1 }× ^{10}C_{6 }**number of ways.

**(e) No girl and 7 boys** will be selected in ** ^{5}C_{0 }× ^{10}C_{7 }**number of ways

**Hence, the number of ways required in this case:**

= ^{5}C_{4 }× ^{10}C_{3 }+ ^{5}C_{3 }× ^{10}C_{4 }+ ^{5}C_{2 }× ^{10}C_{5 }+ ^{5}C_{1 }× ^{10}C_{6 }+ ^{5}C_{0 }× ^{10}C_{7}

=

= 300 + 2100 + 2520 + 1050 + 120

**= 6090 number of ways.**

**Q-4: In a dictionary, the permutation of the letters of the word EXAMINATION is listed, then find the total number of words in the list of dictionary before the first letter of the word start with E.**

** **

**Solution:**

There are **11 letters** in the given word **EXAMINATION** where **A, I and N** appeared **twice** and all other letter appeared only for **once**.

Words which are listed before the word which is starting with **E** in a dictionary can be the word which starts with only **A**.

Hence, in order to get the numbers of **words starting from A**, means the **letter A** will be fixed at the extreme left position of the word, and there are **10** remaining letters taken at a time is rearranged.

In the given word which is **EXAMINATION**, there are **2 N’s and 2 I’s** in the remaining letters and other letters occurred only for once.

Thus, the number of words starting with the letter A:

=

**Hence, the number of words required is 907200.**

**Q-5: Find the total number of 6- digit numbers formed by using the digits 0, 2, 3, 4, 5 and 6 which should be divisible by 10. Note that, the repetition of digit is not allowed.**

** **

**Solution:**

For a number to be **divisible by 10,** then its **unit** **place** must be **filled** with **0 only**.

Thus, **units** place is fixed which is to be filled only **by 0.**

Hence, there are multiple numbers of ways to fill up the other **5 vacant places** in succession with the other digits that is, **2, 3, 4, 5 and 6**.

Now,

Other **5 vacant places** will be filled in by **5!** **number** **of ways.**

**Therefore, the number of 6-digit numbers required = 5! = 120**

**Q-6: It is known that the English alphabet has 21 consonants and 5 vowels. Find the number of words formed with 4 different consonants and 3 different vowels from the English alphabet.**

** **

**Solution:**

We need to select 4 different **consonants** from **21 consonants** and **3 different vowels** from **5 vowels** of the **English** **alphabet**.

So, the **permutation** of the selection of **vowels** is given by:

^{5}**C _{3 }= 5!3!2! = 10**

Also, the permutation of the selection of consonants is given by:

^{21}**C _{2}= 21!4!17! = 3990**

**Hence, the total number of the combinations of 4 different consonants and 3 different consonants = 10 × 3990 = 39900.**

**Q-7: There are 12 questions in the question paper of an examination which is mainly divided into two parts, say, part-I and part-II, each having 4 and 8 questions, respectively. There is a condition given in the question paper that the student has to attempt at least 8 questions for sure, selection 3 from each section. Now, find the number of ways in which a student can select the questions in the question paper.**

** **

**Solution:**

From the data given in the question, know that the **examination** **question** **paper **is **divided** into **two parts**, namely; **part-I and part-II,** containing **4 and 8 questions each.**

A student need to attempt at least **8 questions**, selecting **3 questions** at least from **each** **sections**, which can be done in following ways:

**(a) 4 questions from part I and 4 questions from part II**

**(b) 3 questions from part I and 5 questions from part II**

The selection of **4 questions** **from** **part I** and **4 questions from part II** have permutation as: ^{4}C_{4 }× ^{8}C_{4 }number of ways.

The selection of **3 questions** from **part I** and **5 questions** from **part II** have **permutation** as: ^{4}C_{3 }× ^{8}C_{5 }number of ways.

Now,

**The total number of ways in which selection of questions are required:**

= ^{4}C_{4 }× ^{8}C_{4} + ^{4}C_{3 }× ^{8}C_{5}

=

**= 57 number of ways**

**Q-8: Find the number of ways of 5- card combinations from a deck of 52 cards. Note that, among those 5 cards, there must be exactly one king in the combination.**

** **

**Solution:**

We need to make **5 – card** combinations from a deck of **52 cards**, in such a way that there must be a **king** in the combination.

We know that, there are **4 kings** in a deck of **52 cards**.

Then,

The **permutation** of the selection of a **king** from the **deck** is: ^{4}C_{1 }number of ways.

Now after selection of a **king** for the **combination** of **5 cards** then, there are **4 cards** left to be selected.

Thus, **4 cards** can be selected from the deck with **48 cards** left in ^{48}C_{4}**number of ways.**

**Therefore, the number of 5- card combination required is:**

=** ^{4}C_{1 }× ^{48}C_{4 }**=

**= 7,78,320 number of combinations.**

**Q-9: Consider a situation in which 7 men and 6 women needs to be seated in a row in such a way that the women will occupy the even places.**

** **

**Solution:**

It is required to arrange the seating of **7 men** and **6 women** in a row in such a way that the women will occupy the **even** **places**.

**7 men** can be seated in **7! Ways**. For every arrangement, **6 women** will be seated only at the **blank ( _ ) places** so that the **women** can occupy the **even places.**

A _ A _ A _ A _ A _ A_ A

Thus, the **women** will be **seated** in **6! number** of **ways**.

**Therefore, Number of arrangements possible as per the given condition = 6! × 7! = 3628800 number of arrangements.**

**Q-10: There are 20 students in a class from which 8 students will be chosen for an exclusion party. 3 students from the class decided that either they will join combinely or none of them will go for the exclusion party. Find the number of ways by which the exclusion party will be chosen.**

** **

**Solution:**

**8 students** will be selected for the **exclusion** **party** from a class of **20 students**.

Also, it is given in the question that, **3 students** from that class decided that they will **either** go **together** **or none of them will go for the exclusion party**. Thus, there are mainly 2 cases:

**(a) All 3 of them are going for the exclusion party.**

Thus, the remaining **5 students** will be selected from the remaining **17 students** in the class in **number of ways.**

**(b) None of them are going for the exclusion party.**

Thus, **8 students** will be selected from the class of **20 students** in **number of ways**.

Therefore, the **number of ways** required for **choosing students** for the **exclusion** **party** is:

=

= **6188 + 125970**

**= 132158 number of ways to choose the students for the exclusion party.**

**Q-11: Arrange ASSASSINATION in such a way that all the S’s are filled together. Find the number ways for the arrangement of S’s together.**

** **

Solution:

**ASSASSINATION** is the word given in the question where **A** appeared for **3 times**, **S** appeared for **4 times**,** I** appeared for **2 times**, **N** appeared for **2 times** and, **T and O** appeared just for **once**.

We need to arrange all the words in such a way that the all** S’**s are **placed** together.

Now,

**SSSS** is considered as a **single** **object**.

Since, including the single object we have remaining **9 letters**, so we have the count of **10 letters** more.

Thus, the **10 letters** where they have **3 A’s, 2 I’s and 2 N’s** will be arranged in **number of ways.**

**Therefore, the total number of ways required to arrange the letters of ASSASSINATION word = 10!3!2!2! = 151200**