Question

How many ways can we choose 3 numbers from the set so that more odd numbers are chosen than evan?

Answer 1

Hi there!
Here's the answer:

Choosing from three numbers
=> there are 2³= 8 possible combinations.

say
0- Odd
1- Even

The possible cases are
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1

Here,
we have to choose more odd no.s than even
=> Mark cases in which there are more 0's than 1's
(using (#) symbol for mark)

0 0 0 (#)
0 0 1 (#)
0 1 0 (#)
0 1 1
1 0 0 (#)
1 0 1
1 1 0
1 1 1

Now, count no. of marks gives required no. of possible ways.

Required No. of ways = 4.

:)
Hope it helps