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(A) There is a set where the sequence of numbers x,y,z has n options/possibilities.

(B) x,y,z are in ascending order (between the values of 1 to 4, with 1 and 4 inclusive). x, y and z can also have the same values as one another due to the more than or equal signs.

Basically, I can have x,y,z as 1,2,3 for example, which will fulfill the condition that is set. So how many possible configurations of x,y,z can satisfy this condition?

We simplify this further by asking what happens if I fix x=1?

When x = 1; y,z combinations can be (1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), (4,4) - a total of 10 outcomes.

Repeat this for x = 2, x = 3, x = 4,

x = 1 has 10 outcomes

x = 2 has 6 outcomes

x = 3 has 3 outcomes

x = 4 has 1 outcome

In total, there are 20 outcomes; so n = 20 in this set.

Cheers!