Follows JEE Advanced syllabus, covering these topics:
Difference between Distinguishable and Indistinguishable objects,
Partitioning Indistinguishable Objects,
Counting Number of Groups formed,
No of ways of Choosing objects from a larger set,
Pascal's Triangle and nCr,
Using nCr multiplication to solve Road network problems.
2. Revision of PnC-2
Q1)‘n’th word in an alphabetical ordering of all permutations of a word.
Mechanically find out number of words starting with first letter
onwards, and determine beginning letters of answer word.
Q2)Objects that need to be together in permutations.
‘Tying up’ the objects that need to be together as a single object, and
permuting within them as well.
Q3)Objects that need to be separated in permutations.
Insertion : Permute Objects with no condition as boxes.
Then insert those in between boxes anywhere.
--OR-- Method of Complements : From all permutations, subtract those
where the items are together.
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3. Revision of PnC-2
Q4)’n’ people sitting on a round table.
(n-1)! Permutations. Two methods(Double counting and Fixing 1 Person)
Q5) ’n’ beads in a necklace.
Clockwise-Anti Clockwise symmetry => (n-1)!/2 Permutations.
nCr=nCn-r= n!/(r!(n-r)!)
nCr=n-1Cr+n-1Cr-1
nPr = rPr * nCr= n!/(n-r)!
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4. Solutions to Final Session-2
Q1) If all the letters of the word ARRANGE are arranged in all possible
ways, in how many of words we will have the A's not together?
Answer) Total – A’s are together = 7!/(2!2!) – 6!/2!
Q2) Prove that nCr=n-1Cr+n-1Cr-1 using the formula nCr= n!/(r!(n-r)!)
Answer) Factor out common terms and cancel.
Q3)If nCr=nCr-1, and nPr=nPr+1 then n,r = ?
Answer)
Q4) The value of 50C4 + 56−kC
3= ?
Answer)56C4
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5. Answers to Homework-2
Answer1) 2*3=6
Answer2) 5C1+ 2!5C2+ 3!5C3+ 4! 5C4+ 5!5C5
Answer3) 7C4*3C2=105
Answer4) Total – A’s are together = 6!/2! – 5! = 240
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6. PnC-3 Topics
4.3.Number of ways of Partitioning Distinguishable Objects
4.4.Choosing Arbitrary number of Objects
5.Pascal’s Triangle
5.1.Properties of Pascal’s Triangle
5.2.Proofs on Properties of Pascal’s Triangle
5.3.Road Network Problems
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7. 4.3.Partitioning Distinguishable Objects
Formation of Groups: Number of ways in which (m + n + p) different
things can be divided into three distinct groups containing m, n & p
things respectively
(m≠n≠p)
Method) Original problem is also equivalent to deciding that out of all
permutations of (m + n + p) people, leftmost ‘m’ people will be one
group, middle ‘n’ of another and rightmost ‘p’ of 3rd group.
Hence, number of ways of allocating groups = ?
Recall the example of choosing ‘r’ captains.
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8. 4.3.Formation of Groups
Q)What if m=n=p?
In this case the above method fails.
Hint: Just using the previous answer will end up double counting every
valid sequence.
Reason: All 3 groups could end up consisting of the same people, only
the respective location of the groups in the sequence could be different.
These cases are identical when Grouping people, but still produce
different sequences.
Hence Special case answer = ?
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9. 4.3.Formation of Groups
Example)How many ways of assigning group names Red, Green and
Blue to ‘3m’ people, each group having equal strength.
Solution) In this case the Groups themselves can be distinguished by
their names. Hence answer is same as photo case = ?
When m≠n≠p, Groups can be distinguished by strength.
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10. 4.4.Choosing Arbitrary number of Objects
Type of Problems Discussed :
1. Choosing to buy one or more of ‘n’ distinct items
Application of Multiplication Principle => 2*2*….*2 = 2n
Subtract the single case of empty shopping cart => 2n-1
2. Ways of Choosing one or more of 'p' alike objects of one type, 'q'
alike objects of second type and 'r' alike of third type is:
= (p + 1) (q + 1) (r + 1) – 1
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11. 4.4.Choosing Arbitrary number of Objects
Reason: You buy 0 to ‘p’ items of ‘p’ type. By Addition Principle, (p+1)
options, not 2p because the objects are indistinguishable. Choosing any
fixed number of those is equivalent option, not separate. Now apply
Multiplication.
3. Ways of Choosing one or more of ‘n’ objects of which 'p' alike objects
of one type 'q' alike objects of second type and 'r' alike of third type,
and rest being distinct is:
= (p + 1) (q + 1) (r + 1) 2n - (p + q + r) – 1 (n≥p+q+r)
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12. Problem Session-1(15 min.)
Q1) 1.Number of ways of photographing (2m+n) people of which 2
groups of ‘m’ and a group of ‘n’ people want to stay together = ?
2.Number of ways to divide (2m+n) people into groups of m, m and n
people = ?
Q2) From 5 apples, 4 mangoes & 3 bananas in how many ways we can
select atleast two fruits of each variety if
1.Fruits of same species are identical
2.Fruits of same species are different
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14. 5.1.Properties of Pascal’s Triangle
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1. Each element = sum of 2 elements on top of it.
2. Sum of elements on a horizontal line
= 2n. Can you prove this using induction?
Odd terms terms on
horizontal line gives us 0
4. Sum of elements on diagonal line
gives us Fibonacci sequence.
15. 5.2.Proofs on properties of Pascal’s Triangle
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1. By Method of Construction
2. = nC0+nC1+nC2+….. nCn = n-1C0+ (n-1C0+n-1C1)+(n-1C1+ n-1C2)….. n-1Cn
= 2*(horizontal sum of (n-1) row)
3. Similar to 2.
4. Final Session.
16. 5.3.Road Network Problems
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Example)If only going North or East is allowed ,
How many ways are there from A to B?
Method)This is equivalent to Number of
Permutations of 5 North’s(N’s) and 8 East’s (E’s).
Total objects = 13, 5 of 1 type, 8 of other =>
Answer = ?
17. Final Problem Session(20 min.)
Q1) Prove that each number a in Pascal's
triangle is equal to the sum of the numbers in
the previous left diagonal, starting from the
rightmost number through the number which is
located in the same right diagonal as a.
Q2)Prove that a diagonal sum of elements in
Pascal’s triangle gives us the Fibonacci sequence
<Fi> where Fi= Fi-1+ Fi-2.
Hint for both: use property 1, slide 14. No need
of induction.
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18. Final Problem Session(20 min.)
Q3) How many ways are there to choose two teams of 5 people each
from 15 people?
Use two Methods : Grouping into 3 groups of 5, then choosing, and
Directly choosing(nCr) the teams.
Q4) If r, s, t are prime numbers and p, q are the positive integers such
that the LCM of p, q is r2t4s2, then the number of ordered pairs (p, q)
= ? [IIT – 2006]
Hint: For LCM we take maximum exponent of each prime in
factorization of p and q.
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19. Final Problem Session(15 min.)
Next Lecture we will consider problems like :
If all the letters of the word ARRANGE are arranged in all possible
ways, in how many of words we will have the A's not together as well
as the R's not together ? . Requires the Inclusion Exclusion principle.
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