Find the number of ways to rearrange the letters in the following word. ALABAMA
OK, there are 7 letters and 4 of them are A's three of them are not. Now what?
2007-09-12 07:18:20 · 3 answers · asked by Libby 5 in Science & Mathematics ➔ Mathematics
Were they all different, the answer would have been 7!= 5040. Since 4 A's are the same, you divide this by 4!=24
5040/24=210
2007-09-12 07:40:07 · answer #1 · answered by cidyah 7 · 0⤊ 0⤋
ALABAMA;
there are7 letters with 4 identical letters (A) and the rest all different.The number of permutationsP is given by
7!/4! =7*6*5=210 ANS.
2007-09-12 14:34:57 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Permutation with repetition
P=7!/4! =7*6*5 = 210
In general if you have n elements and a,b,c.. is the number of repeated
P =n!/(a! b! c!..) a+b+c..<=n
2007-09-12 14:39:58 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋