Question 929668: im learning about combinations and permutations and i dont seem to understand how to do this. "how many different areangments can be made using all of the letters in the word HOCKEY" i believe the answer is 720 but i have no way of knowing if it's correct. Another is, "A child has six red blocks and four blue blocks. in how many different ways can the child arrange the 10 blocks in a straight line" i know the answer is 210, but im not sure how to get it. i did 6!•4! but that was incorrect, i also tried go add fhem.
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! For the HOCKEY one:
Six ways to choose the first letter, five ways to choose the next letter, etc...so the number of ways is 6*5*4*3*2*1 = 720.
For the red/blue blocks:
Same as before, we get 10*9*8*7*...*1 = 10!. However this is not correct unless the blocks are all distinguishable. Since we are treating the red blocks the same and the blue blocks the same, we divide by 6! (since there are 6! ways to order the red blocks) and divide by 4! to account for the blue blocks. The number of ways is 10!/6!*4! = 210.
Note: 10!/6!*4! is equal to 10C4 (10 choose 4), and is equal to the number of ways to pick four items out of a set of 10, where order is irrelevant. In this problem, we can essentially pick the four "positions" in the line to assign blue blocks in 10C4 = 210 ways.
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