Permutation & Combination concepts by Gaurav Sharma - Part (2/5)


  • Director, Genius Tutorials, Karnal ( Haryana ) & Delhi | MSc (Mathematics)


    Recommended read: Permutation & Combination concepts by Gaurav Sharma - Part 1

    In how many ways can we select 2 balls from a bag of 5 balls ( 1 Red, 1 Green, 1 Blue, 1 Yellow, 1 Orange) ?

    Answer : 5C2

    But, in how many ways can we select 2 balls from a bag of 5 red balls ?

    since all balls are red in color so answer should be 1

    Number of ways to select r objects out of n distinct objects in nCr

    Number of ways of selecting r objects out of n identical objects is 1

    In how many ways can 5 letters be posted in 4 letter boxes ?

    In this type of questions, one thing is repeatable and the other one is non-repeatable.

    So our answer is always : repeatable ^ non repeatable

    So here the letter boxes are repeatable and the letters are not so our answer is 4^5

    Number of ways to arrange r things out of n when repetition is not allowed are : nPr

    Number of ways to arrange r things out of n when repetition is allowed are : n^r

    Find the number of 3-digit numbers than can be formed using the digits from 1 – 7 when repetition of digits not allowed.

    Answer : Here repetition is not allowed so our answer will be of the form nPr.

    N = 7 and r = 3

    So our answer is 7P3

    Find the number of 3 digit numbers that can be formed using the digits from 1 – 7 when repetition of digits is allowed.

    Answer : Here repetition is allowed so our answer will be of the form n^r

    N = 7 and r = 3

    So, our answer is 7^3

    Number of ways to distribute n identical things to r persons such that anyone can get any number of things is (n+r-1)C(r-1)

    Example : Number of ways to distribute 15 ladoos among 5 persons is (15 + 5 – 1) C (5-1) = 19C4

    If we have to distribute 15 ladoos among 5 persons such that all of them get atleast one ladoo, then we have the equation: A + B + C + D + E = 15

    Now, give one ladoo to each one of them so put A= a + 1, B = b + 1, C = c + 1, D = d + 1 and E = e + 1

    We get, a + b + c + d + e = 10

    Now we can give any number of sweets to anyone of them, so the number of ways in which we can do this is (n + r – 1) C (r – 1)

    N = 10 and r = 5, so (10 + 5 – 1) C ( 5 – 1) = 14C4

    Direct method

    Number of ways of distributing n things to r persons such that all get atleast one are: (n – 1)C(r – 1)

    Positive integral solutions: Where the variables can be any natural number

    Non negative integral solution: Where the variables can be any whole number

    Integral solution: where the variable can be any integer

    Find the number of positive integral solutions of a + b + c + d = 500

    We have to find the POSITIVE integral solutions of a + b + c + d = 500 which means we have to find the natural number of solutions for the equation.

    This case is the same as the case where we have to distribute 500 ladoos among 4 persons such that each gets atleast 1 ladoo.

    So the number of solutions are (n – 1) C (r – 1)

    Here n = 500 and r = 4

    So, (500 – 1) C (4 – 1) = 499 C 3

    Find the non negative integral solutions of a + b + c + d + e .. + z = 500

    We have to find non negative integral solutions for the equations a + b + c + d + .. + z = 500, which means we have to find the whole number solutions for the equation.

    This case is the same as the case where we have to distribute 500 ladoos among 26 persons such that each person can get any number of ladoos.

    So, the number of solutions are (n + r – 1 ) C (r -1 )

    Here, n = 500 and r = 26

    So, (500 + 26 – 1)C(26 – 1) = 525C25

    Find the number of even natural solutions of a + b + c = 100

    a + b + c = 100

    So, put a = 2x, b = 2y and c = 2z

    So 2x + 2y + 2z = 100

    x + y + z = 50

    We need to find even natural solutions of the equation x + y + z = 50

    So the number of solutions = (50 – 1)C(3 – 1) = 49C2

    Find the number of even non negative solutions of a + b + c + d + e = 200

    Put a = 2x, b = 2y and c = 2z, d = 2w and e = 2v

    2x + 2y + 2z + 2w + 2v = 100

    x + y + z + w + v = 100

    So the number of solutions = 100 + 5 – 1 C 5 – 1 = 104C4

    Find the odd non negative integral solutions of  a + b + c  = 159

    Put a = 2x + 1, b = 2y + 1 and c = 2z + 1

    2x + 2y + 2z + 3 = 159

    2 ( x + y + z) = 156

    x + y + z = 78

    Where x, y and z can take values which belong to whole numbers.

    So the number of solutions: (n + r – 1)C(r – 1)

    n = 78 and r = 3

    So, ( 78 + 3 – 1)C(3-1) = 80C2

    Find the number of odd solutions of a + b + c + d = 100

    Put a = 2x + 1, b = 2y + 1, c = 2z + 1 and d = 2v + 1

    2x + 2y + 2z + 2v + 4 = 100

    2 (x + y + z + v) = 96

    x + y + z + v = 48

    Where x, y, z and v can take values which belong to whole numbers because when x = 0 we get a = 1, which is natural number. So the number of solutions: (n + r – 1)C(r – 1)

    Where n = 48 and r =4

    So (48 + 4 – 1)C(4 -1) = 51C3

    Find the number of odd natural solutions of a + b + c + d + e = 200

    Odd numbers added even number of times = Even number
    Odd numbers added odd number of times = Odd number

    So number of solutions for this equation is Zero.

    Find the number of integral solutions of a + b + c + d = 50 such that each variable is greater than 5

    Put a = A + 5, b = B + 5, c = C + 5 and d = D + 5

    We get A + B + C + D = 50 – 5 x 4 = 30

    So, now we have to find positive solutions for

    A + B + C + D = 30

    Which is (n – 1)C(r – 1), where n = 30 and r = 4

    We get, (30 – 1)C(4 – 1) = 29C3

    Find the number of positive solutions of a + b + c = 40 such that they all are distinct

    We need to find Total Solutions – ( Number of solutions where any 2 of the variables are equal + Number of solutions where all 3 variables are equal)

    Total positive solutions = (40 – 1)C(3 – 1) = 39C2

    Number of solutions where any 2 of the variables are equal:

    If a = b we get 2a + c = 40 Where a can take any value from 1 – 19. So, 19 cases.

    If a = 20 then c = 0 which is not considered as a positive solution. Similarly when b = c ( 19 cases ) and c = a ( 19 cases )

    And all three of them cannot be equal because 40 is not a multiple of 3

    So distinct solutions = 39C2 – ( 19 x 3 ) = 684

    Find the number of non-negative integral solutions of a + b + c = 28 such that they all are distinct

    We need to find Total solutions – ( Number of solutions where any 2 of the variables are equal + Number of solutions where all three are equal )

    Total positive solutions = (28 + 3 – 1) C (3 – 1) = 30C2

    Number of the solutions where any 2 of the variables are equal:

    If a = b, we get 2a + c = 28, where a can take any value from 0 – 14. so 15 cases.

    If a = 14 then c = 0 which is considered a non-negative solution.

    Similarly when b = c ( 15 cases) and c = a (15 cases)

    And all three of them cannot be equal because 28 is not a multiple of 3

    So distinct solutions = 30C2 – (15 x 3) = 390

    How many terms are in the expansion of (a + b + c + d)^20

    Sum of powers of each variable in the expansion of (a+b+c+d)^20 will be 20.

    We have to find the number of ways in which 20 can be divided among 4 variables

    So non negative solutions of a + b + c + d = 20

    (n + r – 1)C(r – 1) where n = 20 and r = 4

    S0 ( 20 + 4 – 1)C(4 – 1) = 23C3 = 1771

    Number of terms in the expansion of (a + b + c + d)^20 = 1771

    Find the number of ways of distributing 27 ladoos to Swetabh, Raman and Gaurav such that number of ladoos Swetabg gets are more than the number of ladoos Raman gets which in turn is more than the number of ladoos Gaurav gets

    S + R + G = 27

    S > R > G

    So S, R and G are distinct

    So total solutions – (cases where 2 are equal + cases where all 3 are equal)

    Now total non-negative solutions: (27 + 3 – 1)C(3 – 1) = 29C2

    Number of solutions where any 2 of the variables are equal: If S = R, we get 2S + R = 27 where S can take any value from 0 – 13. So, 14 cases but this includes cases where S = R = G, so 14 – 1 = 13 cases.

    Similarly when R=G (13 cases) and when S = G (13 cases)

    And 1 case where S = R = G = 9

    So distinct solutions = 29C2 – ( 13 x 3 ) – 1 = 366

    And for cases where S > R > G we divide 366 by 3! (Because we want only one of the 6 arrangements possible)

    So answer: 366/6 = 61

     


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