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 a multiple of 3

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

dude   i don't even know. i've been trying for hours.

step-by-step explanation:

c. prove the lengths are the same.

step-by-step explanation:

you need to prove the segments are congruent. congruent segments have the same length. find the lengths of the segments using the distance formula, and compare them.

answer: c. prove the lengths are the same.