Choosing a 5 member team out of 12 girls and 10 boys

We must choose a 5-member team from 12 girls and 10 boys. How many ways are there to make the choice so that there are no more than 3 boys on the team?

The correct answer is (22 choose 5) - (12 choose 1)(10 choose 4) - (10 choose 5).

I understand the (22 choose 5) part, but where I am confused at is the other two parts. I do not know how to get those parts.. Can anyone help me understand how to get to the solution?

-

2 Answers

From the total number of ways to form a $5$ member team we subtract the numbers corresponding to the cases when exactly $4$ boys or exactly $5$ boys are chosen to form teams that have no more than $3$ boys. The first case happens when we choose $4$ boys and $1$ girl and the second happens when we choose $5$ boys and $0$ girls. This gives us $\binom{12+10}{5}-\binom{12}{1}\binom{10}{4}-\binom{12}{0}\binom{10}{5}$.

-

Thank you. I didn't think it would be that simple. – Wooooop Dec 9 '12 at 7:53

That solution proceeds by starting with $\binom{22}5$, the total number of possible $5$-person teams, and subtracting the $\binom{12}1\binom{10}4$ teams that have one girl and four boys and the $\binom{10}5$ teams that have five boys.

The problem could also be solved by noting that there are $\binom{12}2\binom{10}3$ teams with two girls and three boys, $\binom{12}3\binom{10}2$ teams with three girls and two boys, $\binom{12}4\binom{10}1$ teams with four girls and one boy, and $\binom{12}5$ teams with five girls, and calculating the sum

$$\binom{12}2\binom{10}3+\binom{12}3\binom{10}2+\binom{12}4\binom{10}1+\binom{12}5\;,$$

but it’s pretty clearly more efficient to calculate

$$\binom{22}5-\binom{12}1\binom{10}4-\binom{10}5\;.$$

-
I haven't thought about doing it that way! Brian, you have been helping me a lot, and I really appreciate that someone with your knowledge is on here helping others out. Thank you. – Wooooop Dec 9 '12 at 7:57

1
@kevlar: I enjoy it, and you’re very welcome. Quite a few problems of this general type can be attacked either by subtracting bad cases or adding good ones; it’s worth thinking of both possibilities and then deciding which one will involve less calculation. – Brian M. Scott Dec 9 '12 at 8:00
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9590852856636047, "perplexity_flag": "head"}