Answer to Question #178644 in Statistics and Probability for Ku Yi Xi

Question #178644

A word is to be formed using some or all of the 10 letters. M A N C H E S T E R Find the total number of ways of forming the word if a) The word is 10-letter long. The letters can be in any combinations. (2 marks) b) The word is 10-letter long. The word must begin with a consonant and end with a vowel. (5 marks) c) The word is 2-letter long, and must not include T or E. (2 marks) d) The word is 7-letter long. The word must contain at least one E. (6 marks) e) The word is 5-letter long, no other restrictions. (5 marks) 



1
Expert's answer
2021-04-19T17:08:48-0400

We have that:

M A N C H E S T E R

n = 10

a) The word is 10-letter long. The letters can be in any combinations:

"n!=10!=3628800"

b) The word is 10-letter long. The word must begin with a consonant and end with a vowel:

there are 3 vowels and 7 consonants, thus we have 7 possible letters for the begin and 3 possible letters for the end:

"7\\cdot ... \\cdot 3"

Thus we have:

"7\\cdot8\\cdot7\\cdot6\\cdot5\\cdot4\\cdot3\\cdot2\\cdot1\\cdot3 = 846720"

(8 possible letters left for the 2nd position, 7 for the 3rd and so on)

c) The word is 2-letter long, and must not include T or E.

2-letter long word that contains T but does not contain E:

"P(8,2)=\\frac{8!}{2!}=20160"

2-letter long word that contains E but does not contain T:

"P(9,2)=\\frac{9!}{2!}=181440"

Therefore 2-letter long words that do not include T or E:

"20160+181440=201600"

d) The word is 7-letter long. The word must contain at least one E.

7-letter long words:

"P(10,7)=\\frac{10!}{7!}=720"

The 7-letter word does not contain E: thus we have 8 possible letters

"P(8,7)=\\frac{8!}{7!}=8"

Therefore 7-letters words with at least one letter E:

"720 - 8 = 712"

e) The word is 5-letter long, no other restrictions:

"P(10,5)=\\frac{10!}{5!}=30240"


Answer:

a) 3628800

b) 846720

c) 201600

d) 712

e) 30240


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