# How many distinct arrangements of four letters (without repeats) from the set {A,B,C,D,E} are possible?

Again the set is {A,B,C,D,E} and how many arrangment of four without repeats are possible? I'm not entirely sure how to work this problem out. Perhaps there is a formula which could be used? Thanks

• The leftmost letter can be chosen in $5$ ways. For each such choice, the next letter can be chosen in $4$ ways. Continue. – André Nicolas Feb 25 '15 at 5:56
• so we would have 5,4,3,2,1....ok, so then would I add those together or would I multiply them and why? – Diamond Louis XIV Feb 25 '15 at 6:01
• How many $2$-letter words are there? The word can start in $5$ ways, and for everyone of these, there are $4$ ways to continue. So there are $(5)(4)$ $2$-letter (legal) words. Every $2$-letter word can be continued to a $3$-letter word in $3$ ways. So there are $(5)(4)(3)$ $3$-letter words. Almost finished. – André Nicolas Feb 25 '15 at 6:08
• I see now, thank you Andre – Diamond Louis XIV Mar 2 '15 at 0:24
• You are welcome. I imagine you saw that there are $(5)(4)(3)(2)$ such arrangements. – André Nicolas Mar 2 '15 at 5:52

Since an order is implied, we can think about how we would create an arrangement.E.g. $A,B,C,D,E \neq A,B,C,E,D$. So, how to we make a generic arrangement? Well, first note that we have five elements. And, if I wanted to make an arrangement, I need a first element. How many ways can I pick my first element: $5$. But, I can't have the element I picked first again, so I now have $4$ elements to choose from to make a second element in my arrangement. Continue in this way until we have picked the fourth element (lol).
Well, now, the multiplication principle tells us that since I have a task, that I should multiply these numbers together to get $5*4*3*2=5!= {5\choose4}*4!$.