**Q-1: If nC8=nC3, find the value of nC3**

** **

**Solution:**

We know that, if

So, x = y or n = x + y

Hence,

**n = 8 + 3 = 11**

Therefore,

**Therefore,**

**Q-2: Find the value of ‘a’ if:**

**(i) 2aC2:aC2 = 12 : 1 **

**(ii) 2aC2:aC2 = 11 : 1**

** **

**Solution:**

**(i).**

**So, 2a – 1 = 3( a – 2 )**

2a – 1 = 3a – 6

3a – 2a = -1 + 6

**Therefore, the value of a = 5**

**(ii).**

**So, 8a – 4 = 11( a – 2 )**

8a – 4 = 11a – 22

11a – 8a = -4 + 22

3a = 18

**Therefore, the value of a = 6**

** **

**Q-3: Find total number of chords which can be drawn through 21 points on the circle.**

** **

**Solution:**

**2 points** are required to draw one chord on a circle.

Now, in order to obtain the number of chords which can be drawn through the given **21 points** on the circle, we need to count **total number of the combinations**.

Hence, there can be several numbers of chords as there are combinations of **21 points** which can be taken **2 at a time.**

**Therefore, number of chords required = 21C2=21!2!(21−2)!=21!2!19!=21×202 = 210**

**Q-4: Find the number of ways in which a team of 2 boys and 2 girls can be selected from the group of 5 boys and 4 girls.**

** **

**Solution:**

We have to select **2 boys and 2 girls** from a group of **5 boys and 4 girls** to form a team.

Now,

**2 boys** should be selected among **5 boys** group in **ways.**

Also,** 2 girls** should be selected among a group of **4** **girls** in **ways.**

Hence, by the **principle of multiplication**, the number of ways by which a **team of 2 boys and 2 girls** can be formed:

= **= 60 ways**

**Q-5: In how many ways will 9 balls be selected from 7 red balls, 6 blue balls and 5 white balls if every time 3 balls of different colors be selected?**

** **

**Solution:**

We have total of **6 blue balls, 5 white balls and 7 red balls.**

We have to select **9 balls** from these **18 balls** in such a ways that every time **3 different colored balls** can be selected.

Here,

We can select **3 balls** from **7 red balls** in

We can select **3 balls** from **6 blue balls** in

We can select **3 balls** from **5 white balls** in

Therefore, by **the principle of multiplication**, number of ways required to select **9 balls**:

=

**= 7000 ways**

**Q-6: Find out the number of combinations of 5 cards from a deck of 52 cards. Assume that there is at least one ace in each of the combinations.**

** **

**Solution:**

We know that, we have **52 cards** in a deck where there are **4 aces.**

We have to make a combination of **5 cards** in which there will be one ace for sure.

As we have **4 aces** in the deck so, aces can be selected in **number of ways**.

**Therefore**, by the **principle of the multiplication**, number of the combination of **5 cards** required:

=

=

**= 778320 number of combinations.**

**Q-7: Find the number of ways by which one can select a cricket team of 11 players from a bunch of 17 players among which only 5 players can bowl if every cricket team of 11 players must have exactly 4 bowlers.**

** **

**Solution:**

**Among 17 players, 5 players can bowl.**

Now, we need to select a cricket team of **11** **players** where there are exactly **4 players** who **can** **bowl**.

So, Among 5, **4 bowlers** can be selected in **number of ways** and the remaining **7 players** for the team of **11 players** will be selected from the remaining **12 players** in **number of ways.**

Therefore, by the **principle of multiplication**, number of ways required to select a cricket team:

=

=

=

**= 3960 number of ways.**

**Q-8: There are 6 black and 5 red balls. Find the number of ways through which 3 black and 3 red balls will be selected****.**

** **

**Solution:**

We have a bag which contains **6 black** and **5 red balls**.

**3 black balls** will be selected from **6 black balls** in ^{6}C_{3}**number of ways** and **3 red balls** will be selected out of **5 red balls** in ^{5}C_{3 }number of ways.

Therefore, **by** **the principle of multiplication**, required number of ways to select **3 black and 3 red balls**:

=

**= 200 number of ways.**

**Q-9: Find the number of ways by which a student can choose a program of 5 courses, if there are options of 10 courses and for every student 2 courses are made compulsory.**

**Solution:**

**10 courses** are available out of which **2 courses** are made compulsory for each student.

Hence, each student have to choose **3 courses** from the remaining courses, i.e., **8 other remaining courses**, which can be selected in **number of ways.**

Therefore, the total number of ways in which **program **can be chosen:

=

**= 56 number of ways.**