Question

X1 X2 X3 X4 X5 X6 are partitions of a set X such n(X1)= n(X2)= n(X3) and n(X4)= n(X5)= n(X6),
except assume that n(X1)= 8n(X4) and n(X)= 81
Then n(X1)=___________________?
Any clues?

Answer 1

Just let n(X4)=n(X5)=n(X6)=a and then n(X1)=n(X2)=n(X3)=8n(X4)=8a. Since n(X)=n(X1)+n(X2)+n(X3)+n(X4)+n(X5)+n(X6)=81, just plug in variables and solve

Answer 2

What exactly do I plug in
the 8?

Answer 3

You know the number of elements in all of your partitions: \[n(X1)=8a, n(X2)=8a, n(X3)=8a, n(X4)=a,n(X5)=a, n(X6)=a\] You also know that \[n(X)=81=n(X1)+n(X2)+n(X3)+n(X4)+n(X5)+n(X6)\] Plug in the values in the first line into the second line, solve for a to get n(X4), and multiply by 8 to get n(X1), which is what you want to find.

Answer 4

Thank you