**Q-1: Form 3- digit numbers by using digits 4, 5, 6, 7 and 8 and assume that**

**(i) The digits can be repeated.**

**(ii) Digits can’t be repeated in the 3- digit number.**

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**Solution:**

**(i)** There are multiple numbers of ways to form a **3- digit **number by using **5 digits** as there are ways of filling **3 vacant bottles** with **balls** in succession by the given digits.

As per the questions demand, the **repetition** of **digits** is **allowed**. So, the unit places should be filled by any of the **five digits** given in the question.

Similarly,

**Tens and hundreds** places are filled in by any of the **five digits** given in the question.

**Therefore, by the principle of multiplication, by using the given 5 digits i.e., 4, 5, 6, 7, and 8, Total number of 3- digit numbers formed is 5 ****×**** 5 ****×**** 5 = 125.**

**(ii)** As per the questions demand, the **repetition of the digits are not allowed**. Here, if the units place if filled before other places such as tens and hundreds, then it can be filled by any of the five digits i.e., **4, 5, 6, 7 and 8.**

As the digits can’t be repeated and we have **5 digits**, then the total ways of filling the ones place of the **3- digit number is 5.**

Now, remaining numbers that we can use at **tens place is 4**. Now, at tens place, we can use the rest of the 4 digits from the 5 digits given in the question. Similarly, at hundreds place, we can use rest of the 3 digits from the 5- digits given in the question except, that which is already filled at ones place.

By using the **principle** of **multiplication**,

**The total number of ways to form a three- digits numbers without repetition of the given digits is 5 ****×**** 4 ****×**** 3 = 60.**

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**Q-2: Form a 3- digit number which should be even by using 1, 2, 3, 6, 8, and 9. Note that, repetition of digits is allowed. **

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**Solution: **

There are multiple numbers of ways to form a **3- digit number** by using **6 digits**, i.e., **1, 2, 4, 6, 8 and 9,** as the way we fill **3 vacant bottles** with **balls** in succession by the given balls.

The number should be an **even number**. So, the unit places should be filled by either 2 or 6 or 8, i.e., we can fill the **ones** **place** in **3 ways.**

As per the question demand, the **repetition of digits is allowed**. So, the **tens** **place** should be filled by either of the given **6 digits, i.e., 1, 2, 4, 6, 8 and 9** and the **hundreds** **place** should also be filled by either of the **6 digits i.e., 1, 2, 4, 6, 8 and 9** in **6** different ways.

**Therefore, by the principle of multiplication, by using the given 6 digits i.e., i.e., 1, 2, 4, 6, 8 and 9, Total of 3- digit even numbers formed is 3 ****×**** 6 ****×**** 6 = 108.**

**Q-3: Find the total number of 4- letter code formed by using the 10 English alphabets, i.e., a, b, c, d, e, f, g, h, i and j. Assume that the repetition of the letters of English alphabet is not allowed.**

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**Solution:**

There are multiple numbers of ways to form a **4- letter code** ;by using **10 letters** from the English alphabet, i.e., **a, b, c, d, e, f, g, h, i and j,** as the way we fill 4 vacant bottles with balls in succession by the given balls.

**Repetition** of the letters from English alphabet is **not allowed.**

So,

If we start filling up the letters from ones place, then we can fill it in **10** **different** **ways** as we have **10** **letters** from the **English** **alphabets**.

Now, we can fill tens place in **9 different ways** from the remaining **9 letters from the 9 English alphabets**, and also on hundreds place we can fill it by **8 different ways from the remaining 8 letters **from the English alphabets, and at hundreds place, we can fill it in **7 different ways from the remaining 7 letters** from the English alphabets as the **repetition** is **restricted**.

Hence, by the **principle** **of** **multiplication**, by using the given **10** **letters** from the English alphabets, i.e., a, b, c, d, e, f, g, h, i and j, total of 4- letter code formed is 10 × 9 × 8 × 7 = **5040.**

**Therefore, 5040 four- letter code is formed by using the given 10 letters i.e., a, b, c, d, e, f, g, h, i and j from the English alphabets, and there is no repetition in the code hence formed.**

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**Q-4: Find the total number of 5- digit telephone numbers formed by using the numbers from 0 to 9. Note that, each of the telephone number will start with 36 and the repetition of the numbers is not allowed. **

**Solution:**

In the question, it is given that the **telephone** **number** is of 5 **digits** and should start with **36. **

So, there will be several telephone numbers as we have to get **5- digit number** where **2 digit** is fixed that is the first two numbers will be **3 and 6**, respectively, i.e., 3, 6, _, _, _ by using the digits from **0 to 9**, **except** the **digits** **3 and 6**.

If we start filling up the numbers to form a telephone number, from **ones** **place** then we can fill it in **8 different ways** as we have **8 digits from the 0 – 9**.

Now, we can fill **tens** **place** in **7 different ways** from the remaining **7 digits from the 0 to 9**, and also on hundreds place we can fill it in **6 different ways** from the remaining **6 digits** from the **0 to 9**, as the **repetition** is **restricted**.

**Hence, by the principle of multiplication, the total number of ways to form 5- digit telephone number is 8 ****× 7 × 6 = 336.**

**Q-5: Consider a scenario when a person tosses a coin. He tosses the coin for two times and every time the outcome of the toss is recorded by one of his friend. Find the total number of possible outcomes of the coin his friend will record. **

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**Solution:**

A **person** **tossed** a **coin** for **two times** and recorded the outcome every time.

Whenever the **coin** will be **tossed**, the number of **outcome of the coin is 2** which are usually called as a head or a tail or we can say that, in each throw, the different ways to show a **different face is 2.**

**Therefore, by the principle of multiplication, the total number of the outcomes possible by the throw can be 2 ****× 2 = 4.**

**Q-6: Consider 5 flags which are of different colors. Find the total possibility of getting different signals. Note that, each of the signals requires the use of any two flags at the same time, one after the other.**

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**Solution:**

As per the scenario given in the question, **each** of the **signals** uses **2 flags** at a time.

There are so many possibilities of the combination as the number of ways of filling the **2 vacant boxes** in succession of the given **5 flags** which are of **different** **color**.

Now, let us start with the upper portion of the signal be filled with **5 different ways with any of the 5 flags**. After picking up one, the lower portion of the signal will be followed by the upper one, which must be filled up with **4 different signals** by any one of the **4 different colored flags left**.

**Therefore, by the principle of multiplication, the total number of the signals which will be generated by 5 different colors of flag is 5 × 4 = 20.**