How many ways are there to select five unordered elements from a set with three elements when repetition is allowed?
2007-07-22 19:39:49 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Other - Science
I forgot to say, what anser we kept coming up with.
3^5=243, but that is not a correct answer.
2007-07-22 20:08:08 · update #1
Consider that each of the five unordered elements can have any one of three values. If there were only two unordered elements, the number of ways would be 3 times 3, or 3^2. Similarly, for five unordered elements with repetition, the number of ways is 3^5.
2007-07-22 19:53:40 · answer #1 · answered by gp4rts 7 · 1⤊ 0⤋
Why is your answer not correct? Who is telling you it is wrong?
It seems that your base is 3 like a binary number's base is 2 or we count in base 10 (10 fingers) and you have to fill 5 orders.
3 orders of base 10 is 10^3 = 1000, so why not 3^5 = 243.
2007-07-23 17:48:53 · answer #2 · answered by threelegmarmot 2 · 0⤊ 0⤋
Ans.:20
5P3=20
2007-07-22 20:29:05 · answer #3 · answered by zonaith 2 · 0⤊ 0⤋
I hope this is what you mean but here it goes.
Elements:
1A 2B 3C
Order:
A>B>C>B>A
2007-07-22 19:48:56 · answer #4 · answered by pyerzuka 3 · 0⤊ 1⤋