There is not enough information in order to solve
268100. That’s the answer
how many 3 element subsets of {1, 2, 3, 4, 5, 6, 7, 8, ,9, 10, 11} are there for which the sum of the elements in the subset is
AURORKA [14]
Answer:
There are 155 ways in which these elements casn occur.
Step-by-step explanation:
We want 3 element subsets whose sum are multiples of 3
1+2+3= 6
1+2+6= 9
1+2+9= 12
1+9+11=21
1+3+5=9
1+4+8=12
1+5+6=12
1+6+8=15
1+7+10=18
1+8+9=18
1+9+11=21
2+3+7=12
2+4+6=12
2+4+9=15
2+5+11=18
2+6+7=15
2+7+9=18
2+8+5=15
2+8+11=21
2+9+10=21
3+6+9= 18
3+9+11=21
3+10+11=24
6+9+10=27
6+8+11=27
6+7+11=24
7+8+9= 24
8+9+10=27
7+9+11=27 .........
We have 11 elements
We need a combination of 3
The combinations can be in the form
even+ even+ odd
odd+odd+odd
even + odd+odd
So there are 3 ways in which these elements can occur
Total number of combinations with 3 elements =11C3= 165
There are 6 odd numbers and 5 even numbers.
Number of subsets with 3 odd numbers = 6C3= 20
Number of two even numbers and 1 odd number = 5C2*6C1=10*6= 60
Number of 2 odd and 1 even number = 6C2* 5C1= 5*15= 75
So 20+60+75=155
There are 155 ways in which this combination can occur
Answer:
c because it will be half of it meaning you have to multiply rhem
