What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9}for which the sum of those 7 elements is a multiple of 3 ?
(a) 10
(b) 11
(c) 12
(d) 13
(e) 14
(a) 10
(b) 11
(c) 12
(d) 13
(e) 14
The sum of all the 9 elements in the set is 45 which is divisible by 3. So the sum of any 7-element subset of these 9-element set will be divisible by 3 only if the sum of the remaining 2 elements is divisible by 3 too.diebeatsthegmat wrote:What is the number of 7-element subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9}for which the sum of those 7 elements is a multiple of 3 ?
(a) 10
(b) 11
(c) 12
(d) 13
(e) 14
Thus the answer to this question is the same as "in how many ways we can select 2 element from this set such that their sum is divisible by 3?"
The pair of such 2 elements are : (1, 2), (1, 5), (1, 8), (2, 4), (2, 7), (3, 6), (3, 9), (4, 5), (4, 8), (5, 7), (6, 9), (7, 8)
There are 12 such pair. Thus there are 12 such possible 7-element subset
The correct answer is C.
There is also tricky solution to this one.
Note that when a number is divided by 3, only 3 remainders are possible: 0, 1, and 2. Thus the probability that any number randomly selected from a evenly distributed set of integers will be divisible by 3 is (1/3). This fact can be proved. But it's somewhat intuitive.
7-elements can be selected from 9-elements in 9C7 = 36 ways.
Now out of this 36 possible selections, only one-third time the sum of the seven elements will be divisible by 3. Thus number of such 7-element set = (36/3) = 12
Note: This tricky method is applicable here only because the set of given integers consist same number of integers which gives a remainder of 0 or 1 or 2 when divided by 3. If it is the set of first 10 natural numbers, this method will not be applicable then.
