# Thread: Answer check data management

1. ## Answer check data management

46. a) How many five-letter "words" (combinations of letters) can be formed from the word "March"?

Answer: 5! = 5 x 4 x 3 x 2 x 1
= 120 different words can be formed

b) In how many of these words are M and R together?

Answer:
0 letters in front = 2 x 1 x 3 x 2 x 1 = 12
1 letter in front = 3 x 1 x 1 x 2 x 1 = 12
2 letters in front = 3 x 2 x 2 x 1 x 1 = 12
3 letters in front = 3 x 2 x 1 x 2 x 1 = 12

Therefore there are 48 words where M and R will be together.

c) What is the probability of M and R not occurring together?

Answer:
48/120 = x/100
= 40% probability that they will be together

Therefore, there is a 60% probability that M and R won't be together

2. Originally Posted by cnmath16
46. a) How many five-letter "words" (combinations of letters) can be formed from the word "March"?

Answer: 5! = 5 x 4 x 3 x 2 x 1
= 120 different words can be formed

b) In how many of these words are M and R together?

Answer:
0 letters in front = 2 x 1 x 3 x 2 x 1 = 12
1 letter in front = 3 x 1 x 1 x 2 x 1 = 12
2 letters in front = 3 x 2 x 2 x 1 x 1 = 12
3 letters in front = 3 x 2 x 1 x 2 x 1 = 12

Therefore there are 48 words where M and R will be together.

c) What is the probability of M and R not occurring together?

Answer:
48/120 = x/100
= 40% probability that they will be together

Therefore, there is a 60% probability that M and R won't be together
Yes, all very good. Apart from in c), I'd just like to point out that probability of an event is usually expressed as a number between 0 and 1. Not a percentage.

3. Originally Posted by cnmath16
46. a) How many five-letter "words" (combinations of letters) can be formed from the word "March"?

Answer: 5! = 5 x 4 x 3 x 2 x 1
= 120 different words can be formed

b) In how many of these words are M and R together?

Answer:
0 letters in front = 2 x 1 x 3 x 2 x 1 = 12
1 letter in front = 3 x 1 x 1 x 2 x 1 = 12
2 letters in front = 3 x 2 x 2 x 1 x 1 = 12
3 letters in front = 3 x 2 x 1 x 2 x 1 = 12

Therefore there are 48 words where M and R will be together.

c) What is the probability of M and R not occurring together?

Answer:
48/120 = x/100
= 40% probability that they will be together

Therefore, there is a 60% probability that M and R won't be together
(b) can also be done by considering M and R as a single unit and arranging the 4 objects [MR], A, C, H to get 4! = 24.

Then note that within the unit [MR] the M and R can arrange in 2! = 2 ways.

So you get (24)(2) = 48.