in a made up alphabet there are 8 consonants and 4 vowels. How many different arrangements of 5 letters can be made if exactly 2 vowels must be used and no repetition of letters is allowed?
if anyone can tell me how to solve this it would be very much appreciated.
2006-08-27 13:40:39 · 4 answers · asked by nedoglover 4 in Science & Mathematics ➔ Mathematics
That means there are 3 consonants must be used and 2 vowels.
Since there are no repititions, there are (8*7*6)/(1*2*3) possible consonants you can use. There are (4*3)/(1*2) possible vowels you can use. Each selection of letters can be arranged in (5*4*3*2*1 different combinations.
Therefore, there are 40,320 five letters words you can make
(cute♥lil♥azn did not take into account how many ways to distribute the 2 vowels between the 3 consonants. There are ten ways to do that. VVCCC, VCVCC, VCCVC, VCCCV, CVVCC, CVCVC, CVCCV, CCVVC, CCVCV, CCCVV. So, her result needs to be multiplied by 10.)
2006-08-27 13:49:26 · answer #1 · answered by nondescript 7 · 3⤊ 0⤋
well, i think its like this.
5 letters, so you have 5 blanks _*_*_*_*_.
only 2 vowels, so
4*3(no repetition)*_*_*_
again, with no repetition for the consonants, then you have _*_*8*7*6
put it together and you get
4*3*8*7*6
12*56*6
672*6
4032
so thats the answer, if i multiplied right... you might want to check it, just to be sure. hope this helps!
2006-08-27 20:48:12 · answer #2 · answered by ily ♥ 3 · 0⤊ 4⤋
i am not good at math
2006-08-27 21:51:54 · answer #3 · answered by freedom_fighter 2 · 0⤊ 3⤋
do your own homework!
2006-08-27 20:45:53 · answer #4 · answered by Anonymous · 0⤊ 4⤋