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The question is: enter image description here

So after some thinking, my working for part a is

5 and integers less than 5 OR 6 and integers less than 6 OR 7 and integers less than 7

$_4C_3 +_5C_3+_6C_3$

From 4 numbers (number of integers less than 5 in the set), choose 3, etc.

I got 34 selections.

For part b, my working was:

From 4 numbers, choose 2, then from 5 numbers choose 2 (because there are 5 odd numbers to choose from)

$_4C_2 \times _5C_2$

Is my thinking right?

Jim
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1 Answers1

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Part (a) appears to be correct.

In part (b) you've counted the number of selections which have exactly two even numbers, when in fact you want the number of ways which have at least two even numbers. So you need to add in the cases of selections with three even numbers and selections with four even numbers (and these cases are exactly analogous to the one you already did).

Evan Chen
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