*Note: Combinations refer to groups*

You (A), your dad (B) your mom (C) can form how many groups?

Listed below are:

But, are they different groups? No! Why?

Because, each group contains the same members: You three.

Actually, they are different permutations!

Then, how many different groups? Just one!

By repositioning you three, you can’t change your group (combination)!

You can only change permutation (by reordering).

You can change groups only by adding new members (your brother D as in ABCD) or taking off

group members (as in AB from ABC)

So, how do you distinguish Permutation from Combination?

*With respect to order or position*

You change Combinations by adding or pulling out members into or from a Group.

**Now, the Combinations formula: **** **

**For eg,**

** **

**Definition of ****: **

**Taking r things at a time out of n different things, **** groups can be formed****. **

*Groups are easy synonym for Combinations. *

Taking 2 things at a time out of 4 different things, ** **groups can be formed.

Useful Tip:

Follow the tip below to solve combinations and probability problems.

*The number of ways of “selecting” any r things out of n different things is *

The number of ways of “selecting” any 2 things out of 4 different things is

** = 6. **

Interpret 6 () as result (groups) or as a method (ways)

**2. Important Formula on Combinations**

** **

For eg,** **** **

To simplify ** :**

First apply** **** **and bring a smaller number in ‘*r’. *

*So, *

Now, as per 4 in *‘r’*, in the numerator write only product of first four numbers starting from 11

up to 8 and in the denominator write only factorial of 4:

** **

**Find: ** ** **

**Solution**:

** **

**Solved Examples: **

**1. How many cricket teams having 11 players each can be formed out of 16 players? **

Solution: Teams are combinations, in which order has no relevance. Therefore, apply combinations i.e.,

Any 11 out of 16 can be selected in** **** **

Now,

** **

**2. How many committees having 5 persons each can formed out of 5 men and 6 women, if every committee must have 2 men and 3 women?**

**Solution: **

Committees are groups i.e., combinations.

Use the above tip:

Select any 2 men out of 5 in** **** **ways and any 3 women out of 6 in ways.

Each of selecting men and women is an independent task.

Apply *Rule of Counting *to find the total number of ways (groups or combinations):

**× **** ****=**

** **

**3. How many diagonals can be formed by joining six points, no three of which lie on a same line?**

**Solution:**

Join two points to form one straight line.

And join “any” two points to form how many straight lines?

** = 12 straight lines**

But, out of the 12 straight lines, 6 turn into the 6 sides of the hexagon.

Therefore, the remaining 6 straight lines are diagonals.

**Special Formulas on Combinations: **

** + ****+ ****+ ……+ **** = **** **** **

With each group having ** any** number of men, with A, B, C listed below are:

**7 = 23 –1 = 8 –1= 7.**

Similarly, out of *n* different things, taking any number of things, you can form **2n – 1** combinations or groups.

As you need to select at least one,

you will remove **,** from the total list of combinations 2n.

**Solved Example: **

**In a test there are 5 questions. How many ways a student can solve at least one question?**

**Solution**: In 2 ways, a student can handle each question.

Answer it or leave it.

With 5 questions, there are **25** ways of either answering or not answering. Remove the one way in which the student leaves all 5 unanswered.

So, there are **25 –1 = 31** combinations (ways) of answering at least one of 5 questions.