Mathematics Lab Activity-12 Class XI

Mathematics Laboratory Activities on Permutations & Combinations for class XI students Non-Medical. These activities are according to the CBSE syllabus

Chapter - 6

Permutations & Combinations

Activity - 12

Objective

To find the number of ways in which three cards can be selected from given five cards.

Material Required

A sketch pen, a cardboard sheet, white paper sheet, cutter.

Theory

.Fundamental Principal of Counting :

 If an event can occur in m different ways, following which another event can occur in n different ways, then the total number of occurrence of the events in the given order is  m x n.

Factorial Notation : 
The product of n natural numbers is denoted by n!  and read as  n factorial

i.e. n! = 1 . 2 . 3 . 4 .… (n - 2)(n - 1) n  or

      n! = n (n - 1)(n - 2) ……3 . 2 . 1

Permutations:

A permutation is an arrangement in a definite order of  number of objects taken some or all at a time.

In simple words permutation is an arrangement and combination is a selection.

Permutations when repetition is allowed:

The number of permutations of n different objects taken all at a time, when repetition of objects is allowed is  nⁿ

Number of permutations of n different objects taken r at a time, where repetition is allowed is  n^r.

When repetition is not allowed

Permutations when all the objects are distinct.

The number of permutations of n objects taken all at a time, is given by 

The number of permutations of n different objects taken r at a time is given by  

Combinations

It is the method of selecting the objects

The number of combinations of n different objects taken r at a time is given by : 

Procedure

1. Take a cardboard sheet and paste a white paper on it.

2. Cut out 5 identical cards of convenient size from the cardboard, and mark these cards as C₁, C₂, C₃, C₄, and C₅

3. Select one card say C₁ from given 5 cards. Then the other two cards from the remaining four cards can be selected as given below. 

Therefore possible selections of three cards from five are

 {C₁C₂C₃ ,  C₁C₂C₄ ,  C₁C₂C₅ ,  C₁C₃C₄ ,  C₁C₃C₅ ,  C₁C₄C₅ }

4. Select one card say C₂ from given 5 cards. Then the other two cards from the remaining four cards can be selected as given below.

Therefore possible selections of three cards from five are

 {C₂C₁C₃ ,  C₂C₁C₄ ,  C₂C₁C₅ ,  C₂C₃C₄ ,  C₂C₃C₅ ,  C₂C₄C₅ }

5. Select one card say C₃ from given 5 cards. Then the other two cards from the remaining four cards can be selected as given below.

Therefore possible selections of three cards from five are

 {C₃C₁C₂ ,  C₃C₁C₄ ,  C₃C₁C₅ ,  C₃C₂C₄ ,  C₃C₂C₅ ,  C₃C₄C₅ }

6. Select one card say C₄ from given 5 cards. Then the other two cards from the remaining four cards can be selected as given below.

Therefore possible selections of three cards from five are

 {C₄C₁C₂ ,  C₄C₁C₃ ,  C₄C₁C₅ ,  C₄C₂C₃ ,  C₄C₂C₅ ,  C₄C₃C₅ }

7. Select one card say C₅ from given 5 cards. Then the other two cards from the remaining four cards can be selected as given below.

Therefore possible selections of three cards from five are

 {C₅C₁C₂ ,  C₅C₁C₃ ,  C₅C₁C₄ ,  C₅C₂C₃ ,  C₅C₂C₄ ,  C₅C₃C₄ }

8. Count all the possible sections of three cards in steps 3, 4, 5, 6 and 7, we find that there are 30 possible selections and each of the selection repeated thrice. Therefore

The number of distinct relations = 30 ÷ 3 = 10

Observations

1. The number of distinct relations = 30 ÷ 3 = 10

 Number of selections = 10 =  

2. C₁C₂C_{3 ,}  C₃C₁C_{2  ,}  C₂C₁C₃  represents the same relation.

3. C₁C₂C_{4 ,}  C₄C₁C_{2  ,}  C₂C₁C₄   represents the same relation.

4. C₁C₂C_{5 ,}  C₅C₁C_{2  ,}  C₂C₁C₅   represents the same relation.

5. Number of distinct selections of 3 cards from 5 is 10 and it can be calculated by using the formula   

6. Number of distinct selections of 3 cards from 5 is 10 and it can be calculated by using the formula    

 

Result

Number of ways in which three cards can be  selected from five is given by

 

Number of ways in which three cards can be  selected from five is given by

Applications

This activity help us to derive a formula used for combinations of objects

VIVA – VOICE

Q. 1. What is the value of  4!?

Ans. 4!= 4 x 3 x 2 x 1 = 24

Q. 2. What is the value of 15!/13!?

Ans. 210

Q. 3. What is the value of  0!  ?

Ans.  0! = 1

Q. 4. What is the value of   ?

Ans. 

Q. 5. What is the value of  ?

Ans.   

THANKS FOR YOUR VISIT

PLEASE COMMENT BELOW