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

a) The wanted number of ways is equal to the number of permutations of 10 elements with the repetition of two elements (since this word contains two identical letters E).

Then $n = \overline {{P_{8,2}}} = \frac{{10!}}{{2!}} = {\rm{1814400}}$.

Answer: 1814400 ways.

b) 1 consonant letter from a given word can be selected in 7 ways. 1 vowel from a given word can be selected in 3 ways (two vowels are repeated). The remaining 8 letters can be rearranged by 8! ways.

Then

$n = 7 \cdot 2 \cdot 8! = {\rm{564480}}$

Answer: 564480 ways.

c) The first letter can be selected in 7 ways, the second in 6 ways. Then

$n = 7 \cdot 6 = 42$

Answer: 42 ways

d) Find the number of words with one letter E

Since the letter E can be in any of the 7 places, and the remaining 6 letters can be selected in $A_8^6 = \frac{{8!}}{{(8 - 6)!}} = {\rm{20160}}$ ways then number of words with one letter E is equal to

$7 \cdot 20160 =141120$

Find the number of words with two letters E.

Places for the letters E can be chosen in $C_7^2 = \frac{{7!}}{{2!(7 - 2)!}} = 21$ ways, the remaining 5 letters can be selected in $A_8^5 = \frac{{8!}}{{(8 - 5)!}} = {\rm{6720}}$ ways. Then then number of words with 2 letters E is equal to $21 \cdot 6720 = {\rm{141120}}$.

Then total number is 141120+141120=282240.

Answer: 282240 ways.

e) Since two letters E are repeated, the wanted number is

$n = \frac{{A_{10}^5}}{{2!}} = \frac{{10!}}{{2 \cdot (10 - 5)!}} = 15120$

Answer: 15120 ways.