Exercises on Combinaitions and Repetitions

Combinations Dave is the Ceo of a committee. In how many ways can a committee of 5 be chosen from 10 people given, so that Dave ist one of them? a) 252 False! try again. b) 126 Yes! Well done. Explanation c) 495 False! try again. d) 3,024 False! try again. Mixed Exercises Determine whether each of the following situations is a Combination or Permutation. Determining how many different ways you can elect a Chairman and Co-Chairman of a committee if you have 10 people to choose from. Your choice: or Choosing the batting order on a baseball team with 13 players. How many different ways? Your choice: or Choosing 3 toppings for a pizza if there are 9 choices. Your choice: or Voting to allow 10 new members to join a club when there are 25 that would like to join. Your choice: or Creating an access code for a computer site using any 8 alphabet letters. Your choice: or Permutations A special type of password consists of four different letters of the alphabet, where each letter is used only once. How many different possible passwords are there? a) 14,950 False! try again. b) 358,800 Yes! Well done. Explanation c) 426 False! try again. d) 456,976 False! try again. How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive? a) 84 False! try again. b) 120 False! try again. c) 720 Yes! You got it. Explanation d) 504 False! try again. Dave is already chosen prior to the selection, so we need to choose another 4 from 9 remaining (no repetitions allow!). In choosing a committee, order doesn't matter; so we need the number of combinations of 4 people chosen from 9. 9C4 = 9! 4!(9 − 4)!  =  9! 4!·5!  =  9·8·7·6·5! 4!·5!  =  9·8·7·6 4·3·2·1  =  126 The number of permutations of 4 letters chosen from 26 is 26P4 = 26! (26 − 4)!  =  26! 22!  =  26·25·24·23·22! 22!  =  26·25·24·23 = 358,800 The number of permutations of 3 digits chosen from 10 is 10P3 = 10! (10 − 3)!  =  10! 7!  =  10·9·8·7! 7!  =  10·9·8 = 720 Top    |    More Exercises