You are placing cubes that are labeled 1-9 in a row. If the first cube needs to be number 1 and the second needs to be 2, what is the probability that the rest of the cubes will be in order from least to greatest?



Answer :

If the cubes are labeled 1 through 9 and the first cube must be labeled 1, then there are 8 remaining cubes to arrange in ascending order from 2 to 9.

The probability of arranging the cubes in the correct order is the same as the probability of arranging the numbers 2 through 9 in ascending order. Since there is only one correct order out of the total possible orders, the probability is:

\[ \text{Probability} = \frac{1}{n!} \]

where \( n \) is the number of items to arrange. In this case, \( n = 8 \).

So, the probability is:

\[ \text{Probability} = \frac{1}{8!} \]

\[ \text{Probability} = \frac{1}{40320} \]

\[ \text{Probability} \approx 0.0000248 \]

Therefore, the probability that the rest of the cubes will be in order from least to greatest is approximately \( 0.0000248 \), or \( \frac{1}{40320} \).

Hope this helps!

Answer:

  • 1/5040

Step-by-step explanation:

To calculate the probability that the rest of the cubes will be in order from least to greatest after placing the first two cubes (labeled 1 and 2), we need to consider the total number of ways the remaining cubes can be arranged in ascending order.

Given that there are 7 cubes left (3 to 9) to be placed in order, the number of ways they can be arranged in ascending order is 1 (since there is only one way for them to be in order).

Therefore, the probability that the rest of the cubes will be in order from least to greatest is 1 out of the total possible arrangements of the remaining 7 cubes.

The total number of ways to arrange 7 distinct cubes is 7! (7 factorial), which is equal to 5040.

So, the probability is:

  • 1/7! = 1/5040

- Q.E. :))

If u have questions feel free to ask

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