In how many ways can four aces be drawn from a deck of cards? (order is not important.)

There are 52 cards in the deck. there are 4 aces in the deck. you are picking 2 cards out of the deck. at least 1 of them needs to be an ace. assuming exactly 1 of them is an ace, then each ace can be paired with exactly 48 other cards in the deck that are not aces. you have 48 cards that can be paired with the ace of spades. you have 48 cards that can be paired with the ace of clubs. you have 48 cards that can be paired with the ace of hearts. you have 48 cards that can be paired with the ace of diamonds. that's a total of 192 possible pairs. you can also have 2 aces in the pair. this can happen in 6 ways. ace of spades with ace of clubs ace of spades with ace of hearts ace of spades with ace of diamonds ace of clubs with ace of hearts ace of clubs with ace of diamonds ace of hearts with ace of diamonds add 6 to 192 and you get a total of 198 pairs of cards that include at least 1 ace in them. if you look at this in terms of combinations, here's what you have: possible combinations of 2 aces from 4 aces is: c(4,2) = 4!/(2!*2!) = 6 possible combinations of 1 ace and 1 other card is: c(4,1) * c(48,1) = 4 * 48 = 192 add these possible combinations up and you get 198 possible combinations of a pair of cards from a deck of 52 cards where at least one of the4 cards is an ace. the total possible combinations of a pair of cards from a deck of 52 cards is equal to: c(52,2) = 52! / (2!*50!) = 1326. the probability that you will draw a pair of cards with at least 1 ace in there is equal to: 198/1326 = .149321267 this is equal to the number of pairs that have an ace in them divided by the total number of pairs possible.

Answer: r=4.5  feetexplanation: radius  r  is half of the diameter  d,⇒r=d2=92=4.5  feetidk i could be wrong