# Can't understand the solution of this combination

It's a journey with 9 people using two boats.

A boat can have maximum 7 people
B boat can have maximum 4 people


How many ways journeys can be done?

My solution:

A  B
7  2
6  3
5  4


So, number of ways= ${^7C_7}*{^4C_2} + {^7C_6}*{^4C_3} + {^7C_5}*{^4C_4}$ = 12+28+7=47

But in book i see the solution= ${^9C_7} + {^9C_6} + {^9C_5} = {^9C_2} + {^9C_3} + {^9C_4}$ = 246

Why? Can anyone explain how?

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When you say $7 \choose 7$ then you're assuming you're selecting $7$ out of $7,$ not $7$ out of $9.$ When you say $\binom{7}{7} \times \binom{4}{2},$ then you're assuming the order within the boat matters, which is not the case. I.e., $A=(p_1, p_2, \ldots, p_7),B=(\color{red}{p_8}, \color{blue}{p_9})$ is the same as $A=(p_1, p_2, \ldots, p_7),B=(\color{blue}{p_9}, \color{red}{p_8})$. So we don't count them twice.
How many ways can you choose $7$ out of $9$ people to ride boat $A$? If the $9$ people are labelled $p_1, p_2 \dots, p_9,$ then there are $\binom{9}{7}$ ways to do so: $$p_1, p_2, p_3, p_4, p_5, p_6, p_7 \\ p_1, p_2, p_3, p_4, p_5, p_6, p_8 \\ p_1, p_2, p_3, p_4, p_5, p_6, p_9 \\ p_2, p_3, p_4, p_5, p_6, p_7, p_8 \\ \dots$$ Note that every choice above completely defines the choice for $B.$ So the count for the first configuration $(A = 7, B = 2)$ is $9 \choose 7$.
Repeat for the other two configurations $(A = 6,B = 3)$ and $(A = 5, B = 4)$ to get ${9 \choose 7} + {9 \choose 6}+ {9 \choose 5}$.
@newbie are you asking about LaTeX? This website uses MathJax to render LaTeX code. LaTeX is a language for rendering math. You enclose the LaTeX code between . The syntax of LaTeX is standard. You can see this post here and this post. Finally, here is a link to a (slow-loading) page listing MathJax supported commands. –  user2468 Aug 19 '12 at 4:29