Dear Reader,

Below are 4 important types of problems under permutations and combinations topic that **you can expect in IBPS, SBI and RRB exams**. If you know to solve them, you could gain valuable marks that too very easily.

At the end of this simple tutorial, you will find an online practice test.

## Type I: Make A Word Using Letters Given

In this type of problems, you will be given a word. **You have to find the number of words that can be made using the letters in the given word**.

Can’t able to understand this type? Consider a word **“MATHS”**. You can make words like **“AMTHS”**, **“TAMHS”** and a lot more words. **It is not necessary that the new words made have meaning**. In this type of problems, you will be asked to find the total number of words that can be formed.

This type is very easy to solve. Read the below example carefully.

**Example Question 1:** How many words can be formed by using the letters of the word “MATHEMATICAL”?

**Solution:**

**Step 1:**

You have to calculate the number of letters in the given word.

In “MATHEMATICAL” there are 12 letters.

**Step 2:**

You have to count the number of times each letter appears in the given word. You will get the below table

Letter | Number of times the letter occurs |
---|---|

M | 2 |

A | 3 |

T | 2 |

H | 1 |

E | 1 |

I | 1 |

C | 1 |

L | 1 |

**Step 3:**

Now, you can find the number of words that can be made, using the below formula

**Number of words = factorial of total number of letters / product of factorial of repetitions of each letter**

If you substitute the values got in steps 1 and 2, you will get

Number of words = 12! / (2! x 3! x 2! x 1! x 1! x 1! x 1! x 1!)

But, you know 1! = 1. Therefore, above equation becomes

Number of words = 12! / (2! x 3! x 2!)

= (12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (2 x 1 x 3 x 2 x 1 x 2 x 1)

= (12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / 24

= 19958400

## Type II: Make A Word When Some Letters Should Be Together

**This is actually an extension of type I**. This is similar to type I, but additional condition will be given. You will understand this after reading the below example.

**Example Question 2:** Find number of words that can be made from letters of the word “ABACUS” with a condition that “U” and “S” are always together.

**Solution:**

**You could see a condition that “U” and “S” should always be together**.

In a question, when you find a condition that some letters should always be together, **you should treat those letters as 1 (single) letter**.

To solve our example question, you have to assume “U” and “S” as 1 single letter. Therefore, the word can be rewritten as

ABAC(US)

As we already did for example question 1, we have to form a table showing number of times each letter appears. The table is as shown below.

Letter | Number of times the letter occurs |
---|---|

A | 2 |

B | 1 |

C | 1 |

US | 1 |

Note: Here, we have written US as 1 single letter

To find the number of words that can be formed from ABAC(US), you can use the formula (as you saw in example question 1),

Number of words that can be formed from ABAC(US) =

factorial of total number of letters / product of factorial of repetitions of each letter

= 5! / 2!.1!.1!.1!

= 5 x 4 x 3 x 2 x 1 / 2 x 1 x 1 x 1 x 1

= 60 words …. *value 1*

But, this is **not your answer**. There is one more step

You have assumed (US) as 1 (single) letter right? **You have to find number of words that can be formed from the letters “U” and “S”, and multiply this value with answer of previous step**.

If you create repetition table for US, you will get,

Letter | Number of times the letter occurs |
---|---|

U | 1 |

S | 1 |

Number of words that can be formed from U and S = factorial of total number of letters / product of factorial of repetitions of each letter

= 2! / 1!.1!

= 2 …. *value 2*

Now, you have to multiply, value 1 and value 2 to get the answer. Therefore, your answer will be 60 x 2 = **120 words**

*Hint of practice test question:
You already know there is an online practice test at the end of this tutorial. You will see a question of this type in that test. Read and understand this type (II) carefully to solve the practice test question easily.*

## Type III: Combination Problems Based On Selection From Group

This type is very common in bank exams. In this type, **you have to calculate the number of ways a team (short group) can be selected from a large group**.

This type is based on an easy formula, which is given below:

Number of ways a team of R members can be selected from a group of N members is given by **NC _{R}**

*In the above formula, C represent “Combination”.*

The value for NC

The value for NC

_{R}can be calculated as**N! / [R! x (N-R)!]**Below example will help you to understand clearly.

**Example Question 3:** In how many ways can 5 students be selected from a group of 10 students?

**Solutions:**

In the above question, the size of large group N = 10.

From the group of 10 you have to select a team of 5 students.

Therefore, number of members in team = R = 5

You can now calculate the number of ways 5 students can be selected from 10 students using our formula NC_{R}

We have to select 5 students from 10 students.

No of ways of selection = 10C_{5}

= 10! / [(10-5)! x 5!]

= 10! / (5! x 5!)

= (10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / [(5 x 4 x 3 x 2 x 1) x (5 x 4 x 3 x 2 x 1)]

= 252

## Type IV: Problems Based On Selection And Possibilities

**This type is an extension to type 3**. Here, you have to use the same selection from group formula that you saw in type 3. But, in this type, **you have to select from group based on some possibilities (given in question)**. Below example will help you to understand this type clearly.

**Example Question 4:** A box contains 3 red balls, 2 yellow balls and 5 green balls. In how many ways can 3 be drawn from the box, if at least 2 green balls are to be included in the draw?

**Solution:**

You have to select 3 balls. Among the 3 balls at least 2 need to be green.

You can do this in 2 possible ways.

You can select ( all 3 green balls) **OR** ( 2 green balls **AND** 1 non green ball)

In such problems, you have to change **OR** to addition (+) and **AND** to multiplication (X)

Therefore, your final answer will be,

Number of ways of selecting 3 green balls + (Number of ways of selecting 2 green balls **X** Number of ways of selecting 1 non green ball) …. *equation 1*

**Case1: Number of ways of selecting all 3 green balls **

You know that there are 5 green balls. Therefore, size of group, N = 5

You have to select 3 green balls from 5. Therefore, team size, R = 3

Therefore, number of ways 3 green balls can be selected = NC_{R} = 5C_{3}

= 5! / [(5-3)! x 3!]

= 5! / 2!x3!

= 5 x 4 x 3 x 2 x 1 / 2 x 1 x 3 x 2 x 1

= 10

**Case 2: Number of ways of selecting 2 green balls **

In this case, size of group, N = 5 and size of team, R = 2

Therefore, number of ways of selecting 2 green balls = NC_{R} = 5C_{2}

= 5! / [(5-2)! x 2!]

= 5! / 3! x 2!

= 5 x 4 x 3 x 2 x 1 / 3 x 2 x 1 x 2 x 1

= 10

**Case 3: Number of ways of selecting 1 non green ball**

There are 3 red balls and 2 yellow balls. Therefore, there are 5 non green balls in total. Therefore, group size N =5.

You have to select 1 non green ball from 5 non green balls. Therefore, team size, R = 1.

Therefore, number of ways of selecting 1 non green ball = NC_{R} = 5C_{1}

= 5

We simplified 5C_{1} to 5 because **NC _{1} is always equal to N**

*Hint: In combinations, you have to always remember these two rules: NCN = 1 and NC1 = N*

**Final Substitution:**

Now if you substitute the values of cases 1,2 and 3 in equation 1, you will get

Answer = 5C_{3} + (5C_{2} X 5C_{1}) = 10 + (10 x 5)

Answer = 60

Therefore, number of ways of selecting 3 balls so that there are at least 2 green balls = 60 ways.

Ready for short practice test?

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