0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ The leftmost letter can be chosen in $5$ ways. For each such choice, the next letter can be chosen in $4$ ways. Continue. $\endgroup$ – André Nicolas Feb 25 '15 at 5:56
  • $\begingroup$ so we would have 5,4,3,2,1....ok, so then would I add those together or would I multiply them and why? $\endgroup$ – Diamond Louis XIV Feb 25 '15 at 6:01
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Feb 25 '15 at 6:08
  • $\begingroup$ I see now, thank you Andre $\endgroup$ – Diamond Louis XIV Mar 2 '15 at 0:24
  • $\begingroup$ You are welcome. I imagine you saw that there are $(5)(4)(3)(2)$ such arrangements. $\endgroup$ – André Nicolas Mar 2 '15 at 5:52
1
$\begingroup$

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 $5*4*3*2=5!= {5\choose4}*4!$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ he wants 4 out the 5: so choose(5,4) * factoriel(4) $\endgroup$ – r0fg1 Feb 25 '15 at 8:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.