Difference between revisions of "If I have four boxes arranged in a $2 \times 2$ grid, in how many distinct ways can I place the digits $1$, $2$, and $3$ in the boxes, using each digit exactly once, such that each box contains at most one digit? (I only have one of each digit, so one box"
(Created page with "We can think of placing a <math>0</math> in the fourth box that will necessarily be empty. Now the problem is simple: we have four choices of digits for the first box, three f...") |
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We can think of placing a <math>0</math> in the fourth box that will necessarily be empty. Now the problem is simple: we have four choices of digits for the first box, three for the second, two for the third, and one for the last. Thus, there are <math>4\cdot 3\cdot 2\cdot 1 = \boxed{24}</math> distinct ways to fill the boxes. | We can think of placing a <math>0</math> in the fourth box that will necessarily be empty. Now the problem is simple: we have four choices of digits for the first box, three for the second, two for the third, and one for the last. Thus, there are <math>4\cdot 3\cdot 2\cdot 1 = \boxed{24}</math> distinct ways to fill the boxes. | ||
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Latest revision as of 15:24, 30 October 2024
We can think of placing a 
in the fourth box that will necessarily be empty. Now the problem is simple: we have four choices of digits for the first box, three for the second, two for the third, and one for the last. Thus, there are 
distinct ways to fill the boxes.
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