b) In how many of these words are M and R together?
c) What is the probability of M and R not occurring together?
2007-04-15 10:03:01 · 9 answers · asked by Jessie P 1 in Science & Mathematics ➔ Mathematics
a)5!
the letters are different, and position is important
b)2* 4!
we can treat MR as a single lettter, so choosing 4 "letters"
but we could also have RM as a single letter, so its doubled
c) (all-together)/all = (5-2)4!/5! = 3/5
2007-04-15 10:07:30 · answer #1 · answered by hustolemyname 6 · 1⤊ 1⤋
Combinations - just 1
Permutations (order matters)
5 x 4 x 3 x 2 x 1 = 120
2007-04-15 10:07:47 · answer #2 · answered by rosie recipe 7 · 0⤊ 1⤋
a) 5! = 120
b) Consider the following possibility:
MR_ _ _
There are 3! ways the blanks can be filled in.
Now there are 8 ways the MR can be together:
MR _ _ _
_ MR _ _
_ _ MR _
_ _ _ MR
If you reverse the order of the M and R you get 4 more.
So total number of ways = 8 * 3! = 48
c) Number of ways they're not together = 120 - 48 = 72
Prob = 72/120 = 3/5
2007-04-15 10:10:04 · answer #3 · answered by Dr D 7 · 0⤊ 0⤋
a) no. of combinations are =5! =5x4x3x2x1 =120 b)no. of words in which m and r are together=48 c)probability of m and r not occurring together=120--48/120 =0.6
2016-05-20 23:34:41 · answer #4 · answered by marti 3 · 0⤊ 0⤋
Someone needs to do their own homework
2007-04-15 10:07:58 · answer #5 · answered by jmp478 3 · 1⤊ 0⤋
a) 5x4x3x2x1=120
b)
c)
sry, dont no b or c since i took a shortcut
2007-04-15 10:09:16 · answer #6 · answered by Don't Ask 3 · 0⤊ 0⤋
charm
100%
0%
2007-04-15 10:07:59 · answer #7 · answered by Skyhawk 5 · 0⤊ 1⤋
lets see... we can use charm .. umm srry thats all i can c
2007-04-15 10:22:01 · answer #8 · answered by ○•Picasso•○ 5 · 0⤊ 1⤋
I did my homework. You do yours.
2007-04-15 10:08:29 · answer #9 · answered by Alice K 7 · 0⤊ 0⤋