From this thread:
We have a 7-inch board, marked off with six inch-marks.
Choose three of the six inch-marks and cut along those lines.
Hence, there are: \(\displaystyle \displaystyle{6\choose3} \,=\,20\) outcomes.
Here they are: . \(\displaystyle \begin{array}{c}1114 \\ 1141\\ 1411\\4111\end{array}\quad
\begin{array}{c}1222 \\ 2122 \\ 2212 \\ 2221 \end{array}\quad
\begin{array}{c} 1123 \\ 1213 \\ 2113\\1231 \\2131\\2311 \end{array} \quad
\begin{array}{c} 1132\\1312\\3112\\1321\\3121\\3211 \end{array}\)
Calculate how many groups of 4 natural numbes are able to solve this \(\displaystyle x_1+x_2+x_3+x_4\:=\:7\)
Solution is 120 but I don't understand the steps to solve it. . . I don 't agree.
We have a 7-inch board, marked off with six inch-marks.
Code:
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[/SIZE][/SIZE][/SIZE][SIZE=4][SIZE=3][SIZE=3][SIZE=4][SIZE=3][SIZE=3] | : : : : : : |
[/SIZE][/SIZE][/SIZE][/SIZE][/SIZE][/SIZE][SIZE=4][SIZE=3][SIZE=3][SIZE=4][SIZE=3][SIZE=3] | : : : : : : |
[/SIZE][/SIZE][/SIZE][/SIZE][/SIZE][/SIZE][SIZE=4][SIZE=3][SIZE=3][SIZE=4][SIZE=3][SIZE=3][SIZE=4][SIZE=3][SIZE=3] | : : : : : : |[/SIZE][/SIZE][/SIZE]
[/SIZE][/SIZE][/SIZE][/SIZE][SIZE=3][SIZE=3] * - - - - - - - - - - - - - *
[/SIZE][/SIZE][/SIZE][/SIZE]
Choose three of the six inch-marks and cut along those lines.
Hence, there are: \(\displaystyle \displaystyle{6\choose3} \,=\,20\) outcomes.
Here they are: . \(\displaystyle \begin{array}{c}1114 \\ 1141\\ 1411\\4111\end{array}\quad
\begin{array}{c}1222 \\ 2122 \\ 2212 \\ 2221 \end{array}\quad
\begin{array}{c} 1123 \\ 1213 \\ 2113\\1231 \\2131\\2311 \end{array} \quad
\begin{array}{c} 1132\\1312\\3112\\1321\\3121\\3211 \end{array}\)
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