1. How many 10-digit decimal strings are s.t. the sum of their digits modulo 2 e
ID: 3210215 • Letter: 1
Question
1. How many 10-digit decimal strings are s.t. the sum of their digits modulo 2 equals 0?
2. If you knew that the number of 10-digit decimal strings which contain exactly one digit
“5” and exactly one digit “6” is 10·9·88 , then in how many of decimal strings of this type
digit 5 occurs before digit 6? (By “before” I do not mean “right before”, e.g. sequence
(5, 0, 1, 2, 3, 4, 7, 8, 9, 6) satisfies this constraint.)
Please solve these two questions ! Discrete math . Bijection rule can be helpful to solve these problems.
Explanation / Answer
1. As given about 10-digit decimal strings,
Such that sum of their digits modulo 2 equals 0.
With {0,2,4,6,8} as set of Even numbers and {1,3,5,7,9} as set of Odd numbers.
Let us list out all the conditions where, (Sum of the digits) modulo 2 equals '0' ,i.e., the sum of the digits is even.
(i.)When all 10 digits are even => (510) number of strings are possible
(ii.)When all 10 digits are odd => (510) number of strings are possible
(iii.)When 2 digits are odd and 8 digits are even => (52*58) number of strings are possible
(iv.)When 4 digits are odd and 6 digits are even => (54*56) number of strings are possible
(v.)When 6 digits are odd and 4 digits are even => (56*54) number of strings are possible
(vi.)When 8 digits are odd and 2 digits are even => (58*52) number of strings are possible
Hence a total of (6*(510)) number of strings are possible such that the sum of their digits modulo 2 equals 0.
2. Given that if total number of 10-digit decimal strings which contain exactly one '5' and one '6' is 10,988. Then exactly half of them would have 5 before 6, and the other half would have 6 before 5. Hence, number of decimal strings of that type, where digit 5 occurs before digit 6 is (10,988/2) = 5,494.
Hence, 5,494 number of 10-digit decimal strings are present where digit 5 occurs befor digit 6, where both of them are used once.
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