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I have a question about this type of problem solving. the problem I am trying to solve is:

In a group of $6$ boys and $4$ girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

I know how to solve this problem. you calculate all the possible outcomes for when at least one boy is chosen and then add them together. the problem is I don't have an intuitive perception of WHY you add them together. Does anyone have a explanation or a simple example of this type of case?

thank u !

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    Please share your solution to the problem, because the conventional (and simpler) way to solve it is different from what you have stated. We count the total number of ways to choose any $4$ children, and subtract the number of ways to choose $4$ girls: $\binom{6+4}{4}-\binom{4}{4}$. – 2017-01-05
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    ah yes my solution was to count all the "approved" ways of arranging the group. that is the number of ways to choose only one boy + number of ways to choose 2 boys etc. But I can see now that it is simpler to just subtract the ways to choose the 4 girls. – 2017-01-05

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The intuition behind adding different outcomes is the following: Say, that you can conduct an experiment with one way or with another way, then you can conduct the experiment in a total of $1+1=2$ ways. This is exactly the same logic used here.

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Let a simple example you have two coins and asking for possibilities of getting only one head.

We simply said head on first coin and tail on second. But here we are wrong there is a possibility of getting tail on first and head on second.

Then we combine all chances to get what we desired.

Similar case here like you mention first we find with 1 boy, then 2 boys, then 3 and 4 boys. At last add them.

Because at least 1 boy means one boy must. So if we have a group of 4 boys and someone ask you 1 boy present. We answer yes.

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The following is a fundamental principle of counting: If the two sets $A$ and $B$ are disjoint then $|A\cup B|=|A|+|B|$.

For $r\geq0$ denote by ${\cal S}_r$ the set of selections containing exactly $r$ boys. Then the sets ${\cal S}_r$ are disjoint. It follows that the number of admissible selections is given by $\sum_{r=1}^4\bigl|{\cal S}_r\bigr|$.

Remark. Your problem can be solved in a simpler way: Deduct the number of forbidden selections from the number of all selections.