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Session
Permutation Combination - 2
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Session Objective
1. Combination
2. Circular Permutation
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Combination
Combination ? Selection
Selection from a, b, c
a , b , c ,
b , a , c ,
Select one
c , a , b ,
a , b, c ,
b , c , a ,
No. of ways 3
Select two
c , a , b ,
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Combination
Select three
No. of ways 1
Number of selection of some from a group.
Number of rejection of remaining.
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Combination
Number of ways of selecting a group of two
student out of four for a trip to Goa.
S 1, S 2, S 3, S 4
Select two
6 ways.
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Combination
Number of ways of selecting one group Of two for
Goa other for Agra
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Combination
Selection and Arrangement of 3 alphabets from
A, B, C, D.
Selection Arrangement Rejection
A, B, C, ABC, ACB, BAC, BCA, CAB, CBA D
A, B, D, ABD, ADB, BAD, BDA, DAB, DBA C
B, C, D, BCD, BDC, CBD, CDB, DBC, DCB A
A, C, D, ACD, ADC, CAD, CDA, DAC, DCA B
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Combination
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Illustrative Problem
Solution
Man 5 Woman 6 Total - 11
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Solution Cont.
(c) At least 3 man
Man 5 Woman - 6
Composition of Committee Case Man Woman 3 2 5
C3 x 6C2 150 4 1 5C4 x 6C1 30 5 0 5C5
x 6C0 1
No. of Ways 150 30 1 181.
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Illustrative Problem
Solution

• Available persons 5
• Persons to select - 3

Ways 5C3
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Solution Cont.
Case 2 A is not there B may /may not be
there
No. of person - 6
Person to Select - 4
(c) AB never together total no. of
committee - AB always together.
Total no. of committee 6C4 .
AB together in committee 4C2
? No. of Ways 6C4 4C2 9
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Illustrative Problem
Solution
Through two given point and unique straight line
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Illustrative Problem
Find the number of 4 digit numbers that can be
formed by 3 distinct digits among 1,2,3,4,5
No. of digits 5
Solution -
5C3
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Illustrative Problem
In how many ways 9 students can be seated both
sides of a table having 5 seat on each side
(non-distinguishable)
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Circular Permutation
A,B,C,D to be seated in a circular table
Total line arrangement 4!
4 linear Arrangement ? 1 circular arrangement
4 ! linear Arrangement ?
No. of circular arrangement of n object (n-1) !
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Illustrative Problem
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Solution
(i) Boys 5 Girls 4
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Solution Cont.
(ii) Boys 5 Girls 4
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Solution Cont.
(iii) Boys 5 Girls 4
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Invertible Circular Arrangement
Ex - Garland, Neck less.
Clockwise and anticlockwise arrangement
considered as same
For n objects.
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Illustrative Problem
I have ten different color stones. In How many
way I can make a ring of five stones
Solution
Stone - 10
Step 2 - arrange circularly (Invertible)
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Sum of Digits
Find the sum of all three digit numbers formed
by 1, 2 and 3
Sum of digits same (12)
Each digits is repeated same no. of
times 2(3-1) !
All digits comes equal no. of times
Sum of digits in each column (123) x 2! 12
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Sum of Digits
100a 10b c
100x12 10x12 12
12 (102 10 1)
12 x 111
1332
Sum of all numbers
(sum of all digits) x (No. of repetition in a
particular column) x (No. of 1s as number of
digits present in the number
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Class Exercise - 1
In how many ways can 5 boys and5 girls be seated
in a row so that no2 girls are together and at
least2 boys are together?
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Solution
First the boys can be seated in 5p5 5!
ways. Each arrangement will create six gaps __ B
__ B __ B __ B __ B If the girls are seated in
the gaps, no 2 girls will be together. Girls can
be seated in the gaps in 6p5 6! ways. But if
the girls are in the first five or the last five
gaps, no 2 boys will be together. So the girls
can be seated in 6! (5! 5!) 6! 2 5! 4
5!
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Solution contd..
Thus, total arrangements possible are 4 x (5!)2
4 120 120 57600 Note Under the given
condition, more than 2 boys cannot sit together.
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Class Exercise - 2
Find the sum of all numbers formedusing the
digits 0, 2, 4, 7.
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Solution
16 13 4333 208 4333 901264
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Class Exercise - 3
If all the letters of the word SAHARAare
arranged as in the dictionary, whatis the 100th
word?
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Solution
Arranging the letters alphabetically, we have A,
A, A, H, R, S.
Thus, the last word starting with R will be
the100th word. This is clearly RSHAAA.
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Class Exercise - 4
How many numbers can be formedusing the digits
3, 4, 5, 6, 5, 4, 3such that the even digits
occupythe even places?
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Solution
Even digits are 4, 6, 4.
Thus, the total number of ways 3 6 18 ways.
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Class Exercise - 5
Ten couples are to be seated arounda table. In
how many ways can theybe seated so that no two
neighboursare of the same gender?
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Solution
Let all the members of one gender be seated
around the table. This can be done in (10 1)!
ways. Once one gender is seated, arrangement of
other gender is no longer a problem of circular
permutation (since the seats can be identified).
Thus, the second gender can be seated in 10!
ways. Thus, total ways 9! 10!
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Class Exercise - 6
In how many ways can 15 delegatesbe seated
around a pentagonal tablehaving 3 chairs at each
edge?
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Solution
If we consider the problem as one of circular
permutations, the answer is (15 1)! 14! But
we are considering the above two arrangements as
same while they are clearly different. All that
has been done is that all delegates have shifted
one position. One move shift will also give a new
arrangement.
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Solution contd..
Thus, we are counting three different
arrangements as one. Thus, number of actual
arrangements possible 3 14!
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Class Exercise - 7
Prove that the product of r consecutiveintegers
is divisible by r!
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Solution
Let the r consecutive integers be (n 1), (n
2), (n 3), ..., (n r) Product (n 1)(n
2)(n 3) ... (n r)
But nrPr is an integer. Thus, the product of r
consecutive integers is divisible by
r! (Proved)
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Class Exercise - 8
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Solution
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Class Exercise - 9
A person wishes to make up as manyparties as he
can out of his 18 friendssuch that each party
consists of thesame numbers of persons. How
manyfriends should he invite?
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Solution
Let the person invite r friends. This canbe done
in 18Cr ways. To maximise thenumber of parties,
we have to take thelargest value of 18Cr. When n
is even,nCr will be maximum when r n/2. Thus,
he should invite 18/2 9 friends.
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Class Exercise - 10
In how many ways can a cricket teamof 5 batsmen,
3 all-rounders, 2 bowlersand 1 wicket keeper be
selected from19 players including 7 batsmen,6
all-rounders, 3 bowlers and3 wicket keepers?
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Solution
The batsmen can be selected in 7C5 21 ways. The
all-rounders can be selected in 6C3 20
ways. The bowlers can be selected in 3C2 3
ways. The wicketkeeper can be selected in 3C1 3
ways. Thus, the total ways of selecting the
team 21 20 3 3 3780 ways.
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Class Exercise - 11
In how many ways can we select oneor more items
out of a, a, a, b, c, d, e?
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Solution
We can select as in 0 or 1 or 2 or 3, i.e. in 4
ways. We can select b in 2 ways, i.e. either we
select it or we do not select it and so on.
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Class Exercise - 12

• In how many ways can we divide
• 10 persons
• into groups of 5 each,
• (ii) into groups of 4, 4 and 2?

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Solution
We have divided by 2!, because if we interchange
persons in group one, with persons in group two,
the division is not different, i.e. group 1 group
2 group 2 group 1 abdfj ceghi
ceghi abdfj
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