Combinatorics - Possible Solutions for given variables

Combinatorics - Possible Solutions for given variables

• 1.
  Combinatorics Q9
• 2.
  Qn: Combinatorics 2a +5b = 103. How many pairs of positive integer values can a, b take such that a > b? (a) 7 (b) 9 (c) 14 (d) 15
• 3.
  Soln: Combinatorics Let usfind the one pair of values for a, b. a = 4, b = 19 satisfies this equation. 2×4 + 5×19 = 103. Now, if we increase ‘a’ by 5 and decrease ‘b’ by 2 we should get the next set of numbers. We can keep repeating this to get all values. Let us think about why we increase ‘a’ by 5 and decrease b by 2. a = 4, b = 19 works. Let us say, we increase ‘a’ by n, then the increase would be 2n. 2a + 5b = 103. How many pairs of positive integer values can a, b take such that a > b?
• 4.
  Soln: Combinatorics This hasto be offset by a corresponding decrease in b. Let us say we decrease b by ‘m’. This would result in a net drop of 5m. In order for the total to be same, 2n should be equal to 5m. The smallest value of m, n for this to work would be 2, 5. a = 4, b = 19 a = 9, b = 17 a = 14, b = 15 2a + 5b = 103. How many pairs of positive integer values can a, b take such that a > b?
• 5.
  Soln: Combinatorics .. And soon till a = 49, b = 1 We are also told that ‘a’ should be greater than ‘b’, then we have all combinations from (19, 13) … (49, 1). 7 pairs totally. Answer choice (a) 2a + 5b = 103. How many pairs of positive integer values can a, b take such that a > b?