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

Question

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) 

Answer 1

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