### Solved Examples

**Question 1: **Evaluate 18C

_{r} = 18C

_{r + 2 } ** Solution: **

By the property of combination,

we have nC_{x} = nC_{y} => x = y ( or ) x + y = n

18C_{r} = 18C_{r+2 }=> r + r + 2 = 18

=> 2r + 2 = 18

=> 2 r = 18 - 2 = 16

=> r = $\frac{16}{2}$

** r = 8**

**Question 2: **If nP

_{r} = 720 and nC

_{r} = 120, find r.

** Solution: **

We know that nC_{r} = $\frac{nP_{r}}{r!}$

We have nP_{r} = 720 and nC_{r} = 120

Therefore, 120 = $\frac{720}{r!}$ (Substituting in the formula)

=> r! = $\frac{720}{120}$

= 6 = 3 . 2 . 1

=>** r = 3 (**by inspection)

**Question 3: **If thee are 12 persons in a party and if each two of them shake hands with each other, how many handshakes happen in the party?

** Solution: **

We have 12 persons in the party.

It is to note that, when two persons
shake hands, it is counted as one handshake, not two . So this is the
problem on combinations.

The total number of handshakes is same as the number of ways of selecting 2 persons among 12 persons = 12 C_{2}

= $\frac{12.11}{1.2}$

= 11.6

= 66

** There will be 66 handshakes in the party.**

**Question 4: **In how many ways can a cricket team of eleven to be chosen out of a batch of 15 players if,

a. there is no restriction on selection.

b. a particular player is always chosen.

c. a particular player is never chosen.

** Solution: **

There are 15 players and a team of 11 is to be selected.

**a.** Number of ways of selecting 11 players out of 15 is 15C_{11} = 15C_{15 - 11}

= 15C_{4}

= $\frac{15.14.13.12}{1.2.3.4}$

= 1365

**There are 1365 ways of selecting the team with no restriction.**

**b.** If a particular player is is always chosen, this means that 10 players are selected out of the remaining 14 players.

Therefore, required number of ways = 14C_{10} = 14C_{14 - 10}

= 14C_{4}

= $\frac{14.13.12.11}{1.2.3.4}$

= 1001

** There are 1001 ways of selecting the team when a particular player is always chosen.**

c**.** If a particular player is never chosen, this means that 11 players are selected out of the remaining 14 players.

Therefore, required number of ways = 14C_{11} = 14C_{14 - 11}

= 14C_{3}

= $\frac{14.13.12}{1.2.3}$

= 364

** There are 364 ways of selecting the team when a particular player is never chosen.**

**Question 5: **How many triangles can be formed by joining the vertices of a hexagon?

** Solution: **

There are 6 vertices.

One triangle is is formed by selecting a group of 3 vertices from given 6 vertices.

This can be done in 6C_{3} ways.

Therefore, Number of Triangles = 6C_{3} = $\frac{6.5.4}{1.2.3}$

= 5 . 4

= 20

**Twenty triangles can be formed by joining the vertices of a hexagon.**

**Question 6: **In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together.

** Solution: **

Since boys are to be separated.

The 5 girls can be seated in 5 ! ways.

The three boys can be seated in the blanks shown in the following.

** _ G _ G _ G _ G _ G _**

There are 6 blank spaces, therefore, the three boys can be seated in 6C_{3} x 3 ! ways.

Therefore, by the Fundamental Principle of Counting,

the total number of ways = 5 ! x 6C_{3} x 3 !

= 14400 ways

**Therefore there are 14,400 ways of seating 5 girls and 3 boys so that no two boys are together.**