![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.
PnC-3 18](https://image.slidesharecdn.com/pnc3-191007042942/85/Permutations-and-Combinations-IIT-JEE-Olympiad-Lecture-3-18-320.jpg)
Permutations and Combinations IIT JEE+Olympiad Lecture 3
- 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. PnC-3 2
- 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)! PnC-3 3
- 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 PnC-3 4
- 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 PnC-3 5
- 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 PnC-3 6
- 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. PnC-3 7
- 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 = ? PnC-3 8
- 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. PnC-3 9
- 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 PnC-3 10
- 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) PnC-3 11
- 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 PnC-3 12
- 14. 5.1.Properties of Pascal’s Triangle PnC-3 14 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 PnC-3 15 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 PnC-3 16 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. PnC-3 17
- 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. PnC-3 18
- 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. PnC-3 19