Going through a practice exam and have managed to answer all questions except these three, any help is greatly appreciated.
Six people have been given a sheet with a 1 on it and six people have been given a sheet with a 0 on it. If four people are needed for a demonstration of 4 bit numbers:
How many different groups of four people can be formed?
How many different combinations contain two 1's and two 0's?
How many different combinations contain at least two 1's?
You can use the formula for combinations for all of these because the ordering of the group members doesn’t matter,
$$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$
This formula gives you the number of unique groups that can be formed containing $r$ individuals from $n$ individuals.
For part one, you can just apply the formula with $n=12$ and $r = 4$.
For part two you’re more restricted, you’re only going to select two from the ‘1’ group and two from the ‘0’, so you’re going to have two combinations and then multiply those together:
$$\binom{6}{2}\binom{6}{2}$$
For part three, there are a couple ways to do this. You can either start at two and add that to the case when there is three and four. Or you can take all possible groupings and subtract the ones you don’t want. So below I have ‘Total - ((only one 1) + (only 0s))’
$$\binom{12}{4} - (\binom{6}{1}\binom{6}{3} + \binom{6}{0}\binom{6}{4})$$