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CBSE Assignments class 09 Mathematics

  Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX 

Permutation and Combination Class XI Chapter 7

Permutation & Combination Class XI Chapter 7
Explanation of Formulas and basic points related to the permutation and combination class XI chapter 7, method of arrangement and selection of the objects.
Permutation & Combination

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.

This principal can be generalized for any finite number of terms.

  If an event can occur in m different ways, following which another event can occur in n different ways, following which another event can occur in p different ways, and so on.  Then the total number of occurrence of the events in the given order is  m x n x p…………..
Permutation & Combinations

For Example:

A person have 3 pants and 2 shirts. How many different pairs of a pant and a shirt, can he dress up with?

Sol. There are 3 ways to select the pant. He can either select pant P1, P2 or P3. Similarly there are 2 ways to  select  two shirts, S1 or S2
So possible pairs for dress up are 3 x 2 = 6 as shown in the figure
Permutations & Combinations

Example 2
A boy have two bags(B1, B2), 3 tiffin boxes(T1, T2 T3) and 2 water bottles(W1, W2). What is the possible ways to select the three objects.

Sol.
Two bags can be selected in = 2 ways
Three tiffin boxes can be selected in = 3 ways
Two water bottles can be selected in = 2 ways
Total ways of selecting all (2 bags, 3 tiffin boxes, 2 water bottles) = 2 x 3 x 2 = 12 ways
 
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  nr.

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 \[^{n}P_{n} =n!\]The number of permutations of n different objects taken r at a time is given by  \[^{n}P_{r}=\frac{n!}{(n-r)!},\; where\; 0\leq r\leq n\]\[^{n}P_{0}=\frac{n!}{(n-0)!}=\frac{n!}{n!}=1\]\[^{n}P_{n}=\frac{n!}{(n-n)!}=\frac{n!}{0!}=n!\; \because \; 0!=1\]For Example:\[^{7}P_{5}=\frac{7!}{(7-5)!}=\frac{7\times 6\times 5\times 4\times 3\times 2!}{2!}=7\times 6\times 5\times 4\times 3=2520\]\[or\: \: ^{7}P_{5}=7\times 6\times 5\times 4\times 3=2520\]
Derivation of formula \[^{n}P_{r}=\frac{n!}{\left ( n-r \right )!}\]
The number of permutations of n different objects taken r at a time  where 0 ≤ r ≤ n and the objects do not repeat is given by \[^{n}P_{r}=n(n-1)(n-2)(n-3).....(n-r+1)\]\[\because \; \; ^{7}P_{3}=7\times 6\times 5,\: If\: \: n=7\: and \: r=3, \: then\: n-r+1=7-3+1=5\]\[\because \; \; ^{8}P_{3}=8\times 7\times 6,\: If\: \: n=8\: and \: r=3, \: then\: n-r+1=8-3+1=6\]Multiplying numerator and denominator by (n-r)(n-r-1)….3 x 2 x 1, we get\[^{n}P_{r}=\frac{n(n-1)(n-2)(n-3).....(n-r+1)\times (n-r)(n-r-1)....3\times 2\times 1}{(n-r)(n-r-1)....3\times 2\times 1}\]\[^{n}P_{r}=\frac{n!}{(n-r)!}\]
Other Examples
When a coin is tossed then possible outcomes (n) = 2, If we toss a coin for two time (r = 2) then possible arrangements are =  nr  = 22.
When a coin is tossed then possible outcomes (n) = 2, If we toss a coin for three time (r = 3) then possible arrangements are = nr  = 23.
When a die is tossed then possible outcomes (n) = 6, If we toss a die for four time (r = 4) then possible arrangements are = nr  = 64.  

When Repetition is not allowed
Permutations when all the objects are not distinct:
Number of permutations of n objects, where p are of the same kind and rest are all different is given by \[\frac{n!}{p!}\]
The number of permutations of n objects, where P1 objects are of one kind, P2 objects are of second kind ……….. Pk are of kth kind and rest if any are different is given by: \[\frac{n!}{p_{1}!p_{2}!p_{3}!.........p_{k}!},\; \; where\; \; 0\leq r\leq n\]
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 : \[^{n}C_{r}=\frac{n!}{r!(n-r)!}\: \; , \; where\; \; 0\leq r\leq n\]
For Example: Let there are three objects X, Y, Z. We want to find the number of combinations obtained by selecting any two of then. Here we make the selection as given below.
Here order is not important.
So two objects can be selected from 3 in \[^{3}C_{2}\; ways\; =\frac{3!}{2!(3-2)!}=\frac{3\times 2!}{2!\times 1!}=3\]Similarly if there are are seven persons  in a room and each person hand shakes  with each other then total number of hand shakes are given by\[^{7}C_{2}\; ways\; =\frac{7!}{2!(7-2)!}=\frac{7\times 6\times 5!}{2\times 1\times 5!}=21\]Other important results of combinations are:\[^{n}C_{n}=\frac{n!}{n!(n-n)!}=1\]\[^{n}C_{0}=\frac{n!}{0!(n-0)!}=1,\; \;because\; \; 0!=1\]\[If\; \; \; ^{n}C_{a}=^{n}C_{b}\Rightarrow \; either\; \; a=b\; or\; a=n-b\: or\: n=a+b\]\[^{n}C_{r}+^{n}C_{r-1}=^{n+1}C_{r}\]For Example\[^{7}C_{5}=\frac{7!}{5!(7-5)!}=\frac{7\times 6\times 5!}{5!\times 2!}=\frac{7\times 6}{2\times 1}=21\]\[^{7}C_{5}=\: ^{7}C_{7-5}=\: ^{7}C_{2}=\frac{7\times 6}{2\times 1}=21\]
Relation between permutation and combination is given as :\[^{n}P_{r}=^{n}C_{r}\times r!, where 0\leq r\leq n\]
Formula for finding the number of diagonal of a polygon 
If a polygon have n sides then number of diagonals can be calculated by using the formula \[^{n}C_{2}-n\]
Q) If a polygon have 44 diagonals then what is the number of sides of the polygon.
Sol. If a polygon have n sides then number of diagonals are given by \[^{n}C_{2}-n\]According to this question number of diagonals is 44. Therefore \[^{n}C_{2}-n=44\]\[\frac{n(n-1)}{2\times 1}-n=44\]\[n^{2}-3n-88=0\]\[(n-11)(n+8)=0\Rightarrow n=11, n=-8\]But number of sides cannot be negative so rejecting n= - 8. Hence we have n = 11

Q ) How many positive numbers  greater than 6000 and less than 7000 which are divisible by 5 if  no digit is repeated.                                                                         [Ans 112]


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