Data management question: permutations with some identical items?

Question

A coin tossed eight times. In how many different orders could five heads and three tails occur? explain ur resoning.

Answer 1

There are 8! ways to permute 8 objects.

However, in this case, 5 of the objects, the heads, are indistinguishable from one another, and the remaining 3 objects, the tails, are also indistinguishable.

This means that many of the permutations are duplicates. Because there are 5! ways to arrange the heads and 3! ways to arrange the tails, within the 5 spots the head have been assigned to and the 3 spots that the tails occupy, each permutation is one of a set of 5!*3! that are actually identical.

Therefore, you must divide 8! by 5!3! to get the actual number of orders. You end up with 8! / (5!3!) = 40320 / 720 = 56.

Answer 2

Here's my reasoning: With one coin flip, there's two combinations, 2^1 (H, T). With two flips, there's four combinations, 2^2 (HH, HT, TH, HH). With three flips, there's eight combinations, 2^3 (HHH, HHT, HTH, THH, HTT, TTH, THT, TTT). Etc. (So if you want to know, there's 2^8 or 256 ways total for eight tosses.)

Now, going back and making a generalization, in one flip, you can have 1 H and 1 T. In two flips, you have 1 two heads, 2 one head, and 1 with no heads. In three flips, you have 1 three heads, 3 two heads, 3 one heads and 1 with no heads. Any pattern emerging? Pascal's triangle (which incidentally connects with combinations)! For the eighth row, it's 1-8-28-56-70-56-28-8-1. Since we're looking for five heads and three tails, there are 56 ways to have five heads and 3 tails in eight tosses (going from left to right: 1 eight heads, 8 seven heads, 28 six heads, 56 five heads).

Of course, it's easier if you already learned about combinatatorics in high school. Permutations have overlaps, so we use the combination formula: n!/[(n-k)!k!]. Therefore, 8!/(5!*3!) = 40320/720 = 56. That's a little easier in my opinion. On the TI-83+, it's the nCr command.

Answer 3

A permutation is the form approaches you may pass products. In different words, If I easily have 6 billiard balls in six holes, how many approaches am i able to change their positions (with each and each hollow nevertheless containing precisely one ball). i'm no longer somewhat advantageous what advise by way of all possible mixtures, yet ... all possible subsets of a sequence?