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1 Unit Combinatorics State pigeonhole principle. If k pigeons are assigned to n pigeonholes and n < k then there is at least one pigeonhole containing more than one pigeons. Find the recurrence relation satisfying the equation y n A 3 n + B 4 n. y n A 3 n + B 4 n. y n+ A 3 n+ + B 4 n + 3A3 n 4B 4 n y n+ A 3 n+ + B 4 n+ 9A3 n + B 4 n y n+ + y n+ y n 0 Solve a k 3a k, for k, with a 0. Let k0 a k x k a 0 + a x + a x + where is the generating function for the sequence a n. Given a k 3a k 3a k a k 0 Multiplying by x k and summing from to, we have 3 a k x k k 3x k k a k x k a k x k k 0 a k x k 0 3x a 0 + a x + a x + a x + a x + 0 3x a 0 0 [from ] 3x 0 (3x ) 3x k0 a k x k 3 k x k k0 a k Coefficient of x k in G(x) a k 3 n x xn If seven colours are used to paint 50 bicycles, then show that at least 8 bicycles will be the same colour. By Pigeon principle, If there are n pigeons and k holes, then there is at least one hole contains at least k + pigeons. Here k 50, n 7 k n There is at least 8 bicycles will be the same colour. n

2 Solve the recurrence relation y n 8y n + y n 0 for n, where y and y y n 8y n + y n 0 () Let y n r n be the solution of (). r n 8r n + r n 0 r n 8 r + r 0 r n r r 8r + 0 The characteristic equation is r 8r + 0 r 4 0 r 4,4 Hence the solution to this recurrence relation is y n α 4 n + α n4 n y α + α 3 Solving (3) and (4), we get y 3 80 α + 3α α, α 4 Substituting α, α in (), 4 y n 4n + 4 n4n Find the number of non-negative integer solutions of the equation x + x + x 3. If there are r unknowns and their sum is n, then the number of non-negative integer solution for the problem is n + r C r Here there are 3 unknowns and the sum is 7 The number of non-negative integer solutions of the equation x + x + x 3 is + 3 C 3 3C 78 Find the recurrence relation for the Fibonacci sequence. f n f n + f n, n and f 0 0, f Use Mathematical induction show that Let P n : n P : + + P() is true. n k k n n + n + n n + n + ()

3 Let us assume that P(n) is true. Now we have to prove that P(n + ) is true. To prove: P n + : n + n + n + n n + n + n n + n + + n + (from ) n n + n + + n + n + n n + + n + n + n + n + n + n + n + 7n n + n + n + n + n + 3 P n + is true. By induction method, P n : n n n + n + is true for all positive integers. There are 500 students in a college, of these 700 have taken a course in C, 000 have taken a course in Pascal and 550 have taken a course in Networking. Further 750 have taken courses in both C and Pascal. 400 have taken courses in both C and Networking, and 75 have taken courses in both Pascal and Networking. If 00 of these students have taken courses in C, Pascal and Networking. () How many of these 500 students have taken a course in any of these three courses C, Pascal and Networking? () How many of these 500 students have not taken a course in any of these three courses C, Pascal and Networking? Let U denote the number of students in a college. Let A denote the number of students taken a course in C. Let B denote the number of students taken a course in PASCAL. Let C denote the number of students taken a course in Networking. U 500, A 700, B 000, C 550, A B 750, A C 400, B C 75, A B C 00 () The number of students has taken a course in any of these three courses C, Pascal and Networking We know tat A B C A + B + C A B A C B C + A B C A B C 05. () The number of students has not taken a course in any of these three courses C, Pascal and Networking is A B C U A B C Using generating function solve y n+ 5y n+ + y n 0, n 0 with y o and

4 y. Let y n x n () where is the generating function for the sequence y n. Given y n+ 5y n+ + y n 0 Multiplying by x n and summing from 0 to, we have y n+ x n 5 x y n+x n + y n+ x n + y n x n 0 5 x n + y n+ x + y n x n 0 x y x y 0 5 x y [from ] x 5 x + y x y 0 x + 5y 0 x 0 x 5 x + x x + 5 x 0 x 5x + x x 4 x x 5x + x 4x x 4x x 5x + 4x 3x x 4x 3x x A 3x + B x 4x A x + B 3x Put x in () 4 B 3 B B Put x in () A 3 3 A 3 A y n x n 3x x 3x + x 3 n x n + n x n y n Coefficient of x n in G(x) y n 3 n + n+ A box contains six white balls and five red balls. Find the number of ways four balls can be drawn from the box if () They can be any colour () Two must be white and two red

5 (3) They must all be the same colour. Total number of balls + 5 () The number of ways four balls can be drawn from the box if they can be any colour is C ! () The number of ways four balls can be drawn from the box if two must be white and two red C 5C !! (3) The number of ways four balls can be drawn from the box if they must all be the same colour. C 4 + 5C 4 C + 5C ! Prove by the principle of Mathematical induction, for n a positive integer n n n + n + Let P n : n n n + n + () P : + + P() is true. Let us assume that P(n) is true. Now we have to prove that P(n + ) is true. To prove: P n + : n + n + n + n n + n + n n + n + + n + (from ) n n + n + + n + n + n n + + n + n + n + n + n + n + n + 7n n + n + n + n + n + 3 P n + is true. By induction method, P n : n n n + n + is true for all positive integers. Find the number of distinct permutations that can be formed from all the letters of each word () RADAR () UNUSUAL.

6 () The word RADAR contains 5 letters of which A s and R s are there. Te number of possible words 5!!! 30 Number of distinct permutation 30. () The word UNUSUAL contains 7 letters of which 3 U s are there. Te number of possible words 7! 3! 840 Number of distinct permutation 840. Solve the recurrence relation, (n) S(n ) + S(n ), with S(0) 3, S(), by finding its generating function. The given recurrence relation is a n + a n a n 0 with a 0 3, a. Let a n x n a 0 + a x + a x + where is the generating function for the sequence a n. Given a n + a n a n 0 Multiplying by x n and summing from to, we have a n x n + a n x n n x n n n a n x n a n x n + x a n x n n 0 n a n x n 0 x a 0 + a x + a x + + x a x + a x + a x + a 3 x 3 + a 4 x x + x xa 0 + a 0 + a x 0 [from ] x + x 3x x 0 x + x x 3 x 3 x + x x 3 + x x 3 x + x x A + x + B x 3 x A x + B + x Put x in () 3 B + 3 B B 4 3 Put x in () 3 + A + 3A 5 A x + x x x x

7 a n x n 5 n x n a n Coefficient of x n in G(x) n x n x xn a n 5 3 n n Prove, by mathematical induction, that for all n, n 3 + n is a multiple of 3. Let P n : n, n 3 + n is a multiple of 3. () P : is a multiple of 3. P() is true. Let us assume that P(n) is true. Now we have to prove that P(n + ) is true. To prove: P n + : (n + ) 3 + (n + ) is a multiple of 3 (n + ) 3 + n + n 3 + 3n + 3n + + n + n 3 + n + 3n + 3n + 3 n 3 + n + 3 n + n + From () n 3 + n is a multiple of 3 (n + ) 3 + n + is a multiple of 3 P n + is true. By induction method, P n : n, n 3 + n is a multiple of 3, is true for all positive integer n. Using the generating function, solve the difference equation y n+ y n+ y n 0, y, y 0 Let y n x n () where is the generating function for the sequence y n. Given y n+ y n+ y n 0 Multiplying by x n and summing from 0 to, we have y n+ x n x y n+x n + y n+ x n x n + y n+ x y n x n 0 y n x n 0 x y x y 0 x y 0 0 [from ] x x y x y 0 x + y 0 x 0 x x x x + x 0 x x + x x x x x x x x

8 x x x x 3x + x x 3x + x A 3x + B + x x A x + + B 3x Put x in () B B 5 B Put x in () 3 3 A A 5 3 A 3x + + x 3x + x y n x n 3 n x n + n x n y n Coefficient of x n in G(x) y n 3 n + n x xn How many positive integers n can be formed using the digits 3, 4, 4, 5, 5,, 7 if n has to exceed ? The positive integer n exceeds if the first digit is either 5 or or 7. If the first digit is 5 then the remaining six digits are 3,4,4,5,,7. Then the number of positive integers formed by six digits is! 30 [Since 4 appears twice]! If the first digit is then the remaining six digits are 3,4,4,5,5,7. Then the number of positive integers formed by six digits is! 80 [Since 4 & 5 appears twice]!! If the first digit is 7 then the remaining six digits are 3,4,4,5,,5. Then the number of positive integers formed by six digits is! 80 [Since 4 & 5 appears twice]!! The number of positive integers n can be formed using the digits 3,4,4,5,5,,7 if n has to exceed is Find the number of integers between and 50 both inclusive that are divisible by any of the integers,3,5,7. Let A, B, C and D represents the integer from to 50 that are divisible by,3,5 and 7 respectively. A 50 5, B , C , D

9 A B 50 3 A B D B D , A C , A D , B C , C D , A B C , A C D , B C D 50 A B C D The number of integers between and 50 both inclusive that are divisible by any of the integers,3,5,7 is A B C D A + B + C + D A B A C A D B C B D C D + A B C + A B D + A C D + B C D A B C D A B C D A B C D 93 Use mathematical induction to prove the inequality n < n for all positive integer n. Proof: Let P n : n < n () P : < < P() is true. Let us assume that P(n) is true. Now we have to prove that P(n + ) is true. To prove: n + P n + : n + < n < n (from ) n + < n + n + < n + n < n P n + is true. By induction method, n + <. n n + n + < P n : n < n is true for all positive integers. What is the maximum number of students required in a discrete mathematics class to be sure that at least six will receive the same grade if there are five possible grades A, B, C, D and F. By Pigeonhole principle, If there are n holes and k pigeons n k then there is at least one hole contains at least k + pigeons. n Here n 5 k + 5 k 5 k 5 k 5 The maximum number of students required in a discrete mathematics class is. Suppose that there are 9 faculty members in the mathematics department and in the computer science department. How many ways are there to select a committee to develop a discrete mathematics course at a school if the committee is to consist of three faculty member from mathematics department and four from the computer science department?

10 The number of ways to select 3 mathematics faculty members from 9 faculty members is 9C 3 ways. The number of ways to select 4 computer Science faculty members from faculty members is C 4 ways. The number of ways to select a committee to develop a discrete mathematics course at a school if the committee is to consist of three faculty member from mathematics department and four from the computer science department is 9C 3. C 4 ways. 9C 3. C ! ! 770 Using method of generating function to solve the recurrence relation a n + 3a n 4a n 0; n, given that a 0 3 and a. Let a n x n a 0 + a x + a x + where is the generating function for the sequence a n. Given a n + 3a n 4a n 0 Multiplying by x n and summing from to, we have n n a n x n + 3 a n x n 4 n n a n x n + 3x a n x n 4x n a n x n 0 n a n x n 0 a x + a 3 x 3 + a 4 x x a x + a x + 4x a 0 + a x + a x + 0 a 0 a x + 3x 3xa 0 4x 0 [from ] + 3x 4x 3 + x 9x 0 + 3x 4x 3 7x 3 7x + 3x 4x 3 7x + 4x x 3 7x + 4x x A + 4x + B x 3 7x A x + B + 4x Put x in () A A A 9 5 Put x in () 3 7 B + 4 5B 4 B x 5 x

11 a n x n n x n 4 5 a n Coefficient of x n in G(x) x n x xn a n n 4 5 Prove, by mathematical induction, that for all n, n 3 + n is a multiple of 3. Let P n : n, n 3 + n is a multiple of 3. () P : is a multiple of 3. P() is true. Let us assume that P(n) is true. Now we have to prove that P(n + ) is true. To prove: P n + : (n + ) 3 + (n + ) is a multiple of 3 (n + ) 3 + n + n 3 + 3n + 3n + + n + n 3 + n + 3n + 3n + 3 n 3 + n + 3 n + n + From () n 3 + n is a multiple of 3 (n + ) 3 + n + is a multiple of 3 P n + is true. By induction method, P n : n, n 3 + n is a multiple of 3, is true for all positive integer n. Using the generating function, solve the difference equation y n+ y n+ y n 0, y, y 0 Let y n x n () where is the generating function for the sequence y n. Given y n+ y n+ y n 0 Multiplying by x n and summing from 0 to, we have y n+ x n x y n+x n + y n+ x n x n + y n+ x y n x n 0 y n x n 0 x y x y 0 x y 0 0 [from ] x x y x y 0 x + y 0 x 0 x x x x + x 0 x x + x x x x x x x x

12 x x x x 3x + x x 3x + x A 3x + B + x x A x + + B 3x Put x in () B B 5 B Put x in () 3 3 A A 5 3 A 3x + + x 3x + x y n x n 3 n x n + n x n y n Coefficient of x n in G(x) y n 3 n + n x xn Using mathematical induction to show that >, n. 3 n Let P n : > n, n () n P : > > P() is true. Let us assume that P(n) is true. Now we have to prove that P(n + ) is true. To prove: P n + : n + n + > n n + n + n + from () n + n n + + n + n + n + n + > n + n + > n + n +

13 P n + is true. By induction method, n + n + > n + P n : > n, is true for all positive integer n. 3 n Using method of generating function to solve the recurrence relation a n 4a n 4a n + 4 n ; n, given that a 0 and a 8. Let a n x n a 0 + a x + a x + where is the generating function for the sequence a n. Given a n 4a n 4a n + 4 n 4a n 4a n + a n 4 n Multiplying by x n and summing from to, we have 4 a n x n 4 n 4x n n a n x n 4x a n x n + a n x n 4 n n n n n x n a n x n + a n x n 4 n 4x a 0 + a x + a x + 4x a x + a x + + a x + a 3 x 3 + a 4 x 4 + 4x + 4x 3 + 4x a 0 + a x + a x + 4x a 0 + a x + a x + + 4xa 0 + a x + a 3 x 3 + a 4 x 4 + 4x + 4x 3 + 4x 4x + 4xa 0 + a 0 a x x + 4x + 4x + [from ] 4x 4x + + 8x 8x x 4x 4x 4x + x 4x + x x + 4x 4x x x + 8x 4x x + 8x 4x x n x + 8x 4x x A 4x + B x + C x x + 8x A x + B x 4x + C 4x Put x in () x n C C C Put x in () 4

14 + 8 4 A A A 4 4 Put x 0 in () A + B + C 4 + B B 0 4 4x + 0 x x a n x n 4 4 n x n x xn & a n Coefficient of x n in G(x) a n 4 n + n + n + x n + n x n n + xn