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Permutation and Combination short tricks pdf download । Math P&C Short tricks download

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math permutation combination tricks

Permutation and Combination short tricks pdf download । Math P&C Short tricks download

Easy and Smart tricks of Math (permutation and combination) with problem & solution. You can download pdf of these notes, links are given below. Short way to solve math sums are given in theis article.

Parmutation and combination is very important topic of math. Here mycareerstudy.com provides you important notes, tricks and book of this P&C Topic of math.

Take a look at some important rules and tricks of Math (permutation and combination)

General Rule and Formula of (permutation and combination)

PERMUTATION AND COMBINATION METHODS RELATED MATH CALCULATION ARE GIVEN IN BANK EXAMS SO ITS IMPORTANT TO LEARN FOR EXAMS -Math

Factorial Notation
If we consider that ‘n‘ be a positive integer. Then we denoted factorial n as ⌊n or,| n!
n! is defined asn! = n (n-1) (n-2) …………….3.2.1.
Example
5! = ( 1 x 2 x 3 x 4 x 5 )
5! = 120.

4! = ( 1 x 2 x 3 x 4 )
4! = 24.

3! = ( 1 x 2 x 3 )
3! = 6.

Things to Remember
1! = 1 [factorial 1 is always 1]
0! = 1 [factorial 0 is always 1]
Permutation
The different arrangements of a given number or things by taking some or all at a time, are called Permutations.
In Short and basic think, permutation is arrangement, given number or letter, that how we arrange it. we can arrange it taking some number or letter at a time or we can arrange it taking all at a time.

Example
All permutations (or arrangements) made with the letters of a, b, c by taking two at a time are ( ab, ba, ac, ca, bc, cb )
Example
All permutations made with the letters a, b, c, taking all at a time are : ( abc, acb, bac, bca, cab, cba ).
Number of Permutations
Number of all permutations of n things, taken r at a time, is given by:
nPr = n (n-1) (n-2)……(n-r+1)
OR
nPr = n! / (n-r)!

Example
5P2
5! / (5-2)!
= 5! / 3!
= ( 5 x 4 x 3 x 2 x 1 ) / ( 3 x 2 x 1 )
= ( 5 x 4 )
= 20.

Example
8P3
8! / (8-3)!
= 8! / 5!
= ( 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 ) / ( 5 x 4 x 3 x 2 x 1 )
= ( 8 x 7 x 6 )
= 336.

Question 1: In how many ways can the three letters a, b and c be permuted ?

Solution: 
Among all the possible 3 letter permutations of three letters, the first position can have any of the three letters. There are 3 possible ways a letter in the first position can be selected – either a, b or c.

Method of enumerating set of all the ordered arrangements or permutations

So the first place can be occupied by the three letters in 3 possible ways.

For each of these 3 possible arrangements now, the first place has already been decided and we have 2 remaining letters that can be arranged in only 2 distinct ways in the second position.

When we decide the letter for the second position, the letter for the third position is automatically decided as remaining letter is only 1.

Thus, the total number of 3 letter permutations or distinctly ordered arrangements possible for the three letters a, b, c is,

3P3=3×2=factorial(3) =3!3P3=3×2=factorial(3) =3!
=3×2×1=6=3×2×1=6

All possible 3 letter permutations or distinct arrangements of 3 letters a, b, c are,
abc, acb, bca, bac, cab, cbaabc, acb, bca, bac, cab, cba

If there were 4 distinct letters [a,b,c,d][a,b,c,d] instead of three, the total number of different ways to arrange the letters or permutations will be,
4P4=4!=244P4=4!=24

Case 2: Let us now consider another case of all possible permutations of 3 letters out of 4 letters [a,b,c,d][a,b,c,d]. This is permutation of 3 distinct objects out of 4.
Method of enumerating the ordered arrangements or permutations
In this case, first let us fix the first position. We can put any of the 4 letters in the first position. After fixing the 1st position, now we are left with 3 letters and in the second position we can put any of these 3 letters. Lastly after fixing each of the first two positions we will have 2 possibilities for the 3rd position.

So total number of permutations,
4P3=4×3×2=24=4!(4−3)!4P3=4×3×2=24=4!(4−3)!

Similarly, number of permutation of 3 objects out of 5 different objects is,
5P3=5×4×3=60=5!(5−3)!5P3=5×4×3=60=5!(5−3)!

In general, number of permutations of nn distinct objects out of mm distinct objects is,
mPn=m×(m−1)×(m−2)…×(m−n+1)mPn=m×(m−1)×(m−2)…×(m−n+1)
=m!(m−n)!=m!(m−n)!

The expression m×(m−1)×(m−2)…×(m−n+1)m×(m−1)×(m−2)…×(m−n+1) is multiplied and divided by (m−n)!(m−n)! to get m!m! in the numerator and (m−n)!(m−n)! in the denominator.

This is how the well known formula for permutation is derived. But more importantly, it is vital to understand how the number of m×(m−1)×(m−2)…×(m−n+1)m×(m−1)×(m−2)…×(m−n+1) arrangements or permutations are formed following a systematic method.

Take note: Number of permutations of mm out of mm distinct objects is same as (m−1)(m−1) out of mm distinct objects.
mPm=m!(m−m)!=m!0!mPm=m!(m−m)!=m!0!
=m!1!=m![m−(m−1)]!=mPm−1=m!1!=m![m−(m−1)]!=mPm−1

That means 4 out of 4 object permutations, that is 4!=244!=24 is same as 3 out of 4 permutations, which is also 24.

By definition, 0!=10!=1.


Question 2: How many 3 digit numbers can be formed using the digits 1, 2, 3, 4, 5 with no digit repeating?

Solution : The total number of 3 out of 5 permutations of the given digits are,
5P3=5!(5−3)!5P3=5!(5−3)!
=5!2!=5×4×3=60=5!2!=5×4×3=60

Questions 3: How many 3 digit numbers greater than 300 can be formed using the digits 1, 2, 3, 4, 5 and 6 with no digit repeating?

Solution: Here the possible digits in the first position of the resultant numbers will be 3, 4, 5 or 6. The digits 1 or 2 in the first position will make the resultant number less than 300.
After deciding the digit in the first position in 4 possible ways we are left with 2 positions and 5 digits. So the desired number of permutations is,
4×5P2=4×5!(5−2)!4×5P2=4×5!(5−2)!
=4×5!3!=4×5×4=80=4×5!3!=4×5×4=80


Question 4: How many two digit numbers can be formed out of the digits 2, 3, 4 and 0 without repeating digits?


Solution 4: The desired number of permutations may at first seem to be a 2 out of 4 permutation, which is 12. But if you are careful you would remember that in the first position of a number (which is the most significant position) placing a 0 is ignored. If we place a 0 in the first position, the resultant number will then become a single digit number.
So we can place only 3 digits 2, 3 and 4 in the first position. Rest of the digits are then 3 and number of positions 1. This second position can then be filled in 3 ways.
So the desired number of numbers is 3×3=93×3=9 instead of 12.
It will be 3 less than 12 for the prohibition of 0 in the first position.
Important note:
If you are clear about the concept of how ordered arrangements or permutations are formed, you should always be able to approach a problem of permutation in this way and should use the formula of permutation only when its application is absolutely straightforward without any complication.

As the formula of permutation comes only after you enumerate the permutations systematically and exhaustively following this method in general, the method forms the basic concept layer that is to be understood very clearly. The formula is trivial and will follow immediately out of the results of the method.

A big advantage of the systematic method of enumeration of ordered arrangements is its ability to generate the actual arrangements along with the number of arrangements, while the formula would give you only the number of arrangements but not the actual permuted arrangements.

Permutation of objects that are not distinct
In this case, we have some of the objects that are alike. Let us consider a mm out of mm permutation in which pp objects are alike. We must form the arrangements that are distinct considering the order of the objects. Otherwise an arrangement will not be a distinct permutation.

Let us assume that the desired number of permutations is xx. In each of these permutations, there are pp objects in various positions of the arrangements. Let us take one of these xx arrangements. In this arrangement, if we make these pp objects all distinct now, we will have p!p! ways to rearrange these now different pp objects among the particular positions of the specific arrangement of xxchosen.

Thus if we make the pp objects distinct, for each of the xx arrangements we will have p!p! new arrangements and we will reach the mm out of mm distinct object permutations. This means,
mPm=m!=x×p!or, x=m!p!mPm=m!=x×p!or, x=m!p!

Ultimately it is a simple relationship – for pp alike objects in mm objects, the mm out of mm permutation is m!m! divided by p!p!.

If there are qq more alike objects in mm in addition to pp alike objects, we just need to divide our previous permutation by q!q!. Thus, for pp, qq and rr alike objects in mm objects the number of mm out of mm permutations is,
m!p!q!r!m!p!q!r!


Questions 5: How many 6 digit numbers can you form from the digits 1, 2, 3, 4, and 5 with 1 repeating in each number twice?

Solution:
The desired number would be 6 out of 6 permutations from the digits, 1, 1, 2, 3, 4 and 5. Here 1 is repeated twice. So the desired number of permutations is,
6!2!=6×5×4×3×2×12×1=3606!2!=6×5×4×3×2×12×1=360


Question 6: In how many ways a person can invite his 6 friends by sending invitation cards through 2 of his servants?

Solution:
This is a different type of problem from permutation. In this case, the person is free to send any of the two servants to each of the six friends. Thus, each friend can receive his card in two possible ways from two servants. As there are six friends, the required number is,
2×2×2×2×2×2=26=642×2×2×2×2×2=26=64

Combination

Basic concepts

Combination is selection of rr objects from a set of nn distinct objects. In this case, there is no importance attached to the order of selection. Each unique set of rr objects selected form one combination.

As in each such combination the rr objects selected can be ordered among themselves in r!r!unique ways, if we order all the combinations in this way we would get the permutation of rr out of nn distinct objects. Thus, number of combinations multiplied with r!r! gives us number of permutations.
So,
Number of combinations,nCr=Number of Permutationsr!nCr=Number of Permutationsr!
=n!r!(n−r)!=n!r!(n−r)!

Question 1: In how many ways can you select 3 books out of 5 available books?

Solution:
The number of ways 3 books can be selected out of 5 books is,
5C3=5!3!(5−3)!=5!3!2!=105C3=5!3!(5−3)!=5!3!2!=10


Questions 2: In how many ways 4 members can be selected out of 8 members to form a committee such that 1 member is always selected?

Solution:
If 1 member is always selected in the committee then the combination choice problem is changed to selecting (4−1)=3(4−1)=3 members out of (8−1)=7(8−1)=7 members. The required number of ways then,
7C3=7!3!(7−3)!=7!3!4!=357C3=7!3!(7−3)!=7!3!4!=35

Question 3: In how many ways 4 members can be selected out of 8 members to form a committee such that 2 members are always excluded?

Solution:
If two members are always excluded, the number of members to choose from reduces to (8−2)=6(8−2)=6 and the required number of combinations is,
6C4=6!4!(6−4)!=6!4!2!=156C4=6!4!(6−4)!=6!4!2!=15

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