Permutation of the eight letters (i, l, o, v, e, y, o, u)

Question

Determine how many permutations of the eight letters (i, l, o, v, e, y, o, u) begin with you, or end with i, or have the letter e in the fifth position and the letter y in the sixth position.

Though it's very easy question, I need to double check the solution.

My solution:

i) you_ _ _ _ _ = 5!

ii) _ _ _ _ _ _ _ i = 7!/2!

iii) _ _ _ _ e y _ _ = 6!/2!

Removing the duplicates:

ii) you _ _ _ _i = 4!

iii) _ _ _ _ e y _ i = 5!/2!

So, total number of permutations = 5! + 7!/2! + 6!/2! - 4! - 5!/2!

Please correct me if I am wrong.

Answer 1

My solution:

i) you_ _ _ _ _ = 5!

ii) _ _ _ _ _ _ _ i = 7!/2!

iii) _ _ _ _ e y _ _ = 6!/2!

Removing the duplicates:

ii) you _ _ _ _i = 4!

iii) _ _ _ _ e y _ i = 5!/2!

So, total number of permutations = 5! + 7!/2! + 6!/2! - 4! - 5!/2!

You are correct.

Answer 2

Yes, that looks correct.

You have accounted each case correctly, and applied the principle of inclusion and exclusion.

• $A$ : begin with "you", 5 letters left to permute, incl. 1 "o"
• $B$ : end with "i", 7 letters left to permute incl. 2 "o"
• $C$ : have the letter "e" in the fifth position and the letter "y" in the sixth position, 6 letters left to permute incl. 2 "o"
• $A\cap B$, 4 letters left to permute, incl. 1 "o"
• $B\cap C$, 5 letters left to permute, incl. 2 "o"
• $A\cap C$, and $A\cap B\cap C$ not possible, as only one "y" is available. $${\quad\mathsf P(A\cup B\cup C)\\=\mathsf P(A)+\mathsf P(B)+\mathsf P(C)-\mathsf P(A B)-\mathsf P(BC)-\mathsf P(A C)+\mathsf P(ABC)\\=5!+7!/2!+6!/2!-4!-5!/2!-0+0\\=2\,916}$$