Mathematics. How Many Different Ways Can You Rearrange 9 Items So That There Are No Duplicate Arrangements?

Mathematics : How Many Different Ways Can You Rearrange 9 Items So That There Are No Duplicate Arrangements

This is not a factorial because in this problem each item can only be used once. So 111111111 is not a viable option, nor would be both 12 and 21 because they are the same items just in different order. I need to know all of the possible arrangements ranging from 1 unit in length all the way to 9 units in length. ~~~ andazzo ~~~

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I presume you mean, say, the first 9 letters of the alphabet. Starting with a choice of 1 item, there are obviously 9 ways. Then with 2 items, you have 9C2 ways. ., factorial 9 divided by (factorial 7 times factorial 2) which is (9 X 8 X ---X 4 X 3 X 2 X 1)/ [(7 X 6--- X 2 X 1)( 2 X 1)] This simplifies to 9 X 8/(2 X 1) = 36. nCr means factorial n divided by (factorial r X factorial (n-r)) In shorthand notation, factorial k is expressed as k! Note that (n C r) gives the COMBINATIONS and not the PERMUTATIONS, so there are none of the duplicates you mentioned. Do the same for 9C3, 9C 4 etc, and go all the way to 9C9. Once youve done that, add up the 9 results to get your answer. You said that this is not a factorial. In fact it is. For example, lets see how many ways you can choose 2 letters out of the first 4 letters of the alphabet (no duplicates). The answer is 4C2 = 4!/(2! X 2!) = 4 X 3 X 2 X 1/ (2 X 1 X 2 X 1) = 6 Here they are: AB AC AD BC BD CD Hope this helps

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Answer 1

I presume you mean, say, the first 9 letters of the alphabet. Starting with a choice of 1 item, there are obviously 9 ways. Then with 2 items, you have 9C2 ways. ., factorial 9 divided by (factorial 7 times factorial 2) which is (9 X 8 X ---X 4 X 3 X 2 X 1)/ [(7 X 6--- X 2 X 1)( 2 X 1)] This simplifies to 9 X 8/(2 X 1) = 36. nCr means factorial n divided by (factorial r X factorial (n-r)) In shorthand notation, factorial k is expressed as k! Note that (n C r) gives the COMBINATIONS and not the PERMUTATIONS, so there are none of the duplicates you mentioned. Do the same for 9C3, 9C 4 etc, and go all the way to 9C9. Once youve done that, add up the 9 results to get your answer. You said that this is not a factorial. In fact it is. For example, lets see how many ways you can choose 2 letters out of the first 4 letters of the alphabet (no duplicates). The answer is 4C2 = 4!/(2! X 2!) = 4 X 3 X 2 X 1/ (2 X 1 X 2 X 1) = 6 Here they are: AB AC AD BC BD CD Hope this helps