For the captain to be seated at one end of the rectangular table: - There are 2 choices for which end the captain sits at. - There are 9! ways to arrange the remaining people on the other seats. Hence, the total number of ways is \( 2 \times 9! \). <p>\( 2 \times 9! = 2 \times 362,880 = 725,760 \)</p>
The image you've provided is somewhat blurry, but I can make out a question asking to fill in a blank. It asks about lotto with n choices from a collection of m numbers (n < m), and how many different outcomes (selections) are possible. The question seems to be about combinations, where the order in which the numbers are drawn does not matter. The formula to calculate combinations is given by: \[ {m \choose n} = \frac{m!}{n!(m-n)!} \] Here m! represents the factorial of m, which is m multiplied by all positive integers less than m, n! is the factorial of n, and (m-n)! is the factorial of the difference between m and n. Plug in the values of m and n into this formula to find the number of different lottery outcomes that are possible. The values of m and n, however, are not visible in the image, so you would need to provide those specific numbers in order to calculate the exact number of outcomes. If you can provide those numbers or a clearer image, I would be glad to assist with the calculation.
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