Complete question :
How many different pairs (m, n) can be formed using numbers from the list of whole numbers { 1, 2, 3, ..., 20 } such that m < n and m + n is even?
Answer:
90
Step-by-step explanation:
Given :
Condition :
m < n ; m + n = even number
Thus ;
m = 1 ; n = 3, 5, 7, 9, 11, 13, 15, 17, 19 = 9
m = 2; n = 4, 6, 8, 10, 12, 14, 16, 18, 20 = 9
m = 3 ; n = 5, 7, 9, 11, 13, 15, 17, 19 = 8
m = 4 ; n = 6, 8, 10, 12, 14, 16, 18, 20 = 8
m = 5 ; n = 7, 9, 11, 13, 15, 17, 19 = 7
m = 6 ; n = 8, 10, 12, 14, 16, 18, 20 = 7
Observing the trend :
For ;
m = 7, number of ways = 6
m = 8, number of ways = 6
m = 9, number of ways = 5
m = 10, number of ways = 5
m = 11, number of ways = 4
m = 12, number of ways = 4
m = 13, number of ways = 3
m = 14, number of ways = 3
m = 15, number of ways = 2
m = 16, number of ways = 2
m = 17, number of ways = 1
m = 18, number of ways = 1
m = 19, Number of ways = 0
Hence,
(9+9+8+8+7+7+6+6+5+5+4+4+3+3+2+2+1+1) =90
Hence, we have 90 (m, n) pairs
