4
$\begingroup$

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.

$\endgroup$

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Perhaps it looks like a joke but I did check your work! (: $\endgroup$
    – Landon
    Commented Nov 8, 2019 at 4:29
  • $\begingroup$ I reckon that 'I checked your work. You are correct.' will suffice here. It might look odd, but if it answers the question it will be fine. However, if no one comes up with another answer detailing how to get to the solution, then you should add another answer, since we want questions to be useful to more than just the OP and the answerers. $\endgroup$
    – Toby Mak
    Commented Nov 8, 2019 at 13:08
4
$\begingroup$

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. $${\quad\mathsf P(A\cup B\cup C)\\=\mathsf P(A)+\mathsf P(B)+\mathsf P(C)-\mathsf P(A B)-\mathsf P(BC)-\mathsf P(A C)+\mathsf P(ABC)\\=5!+7!/2!+6!/2!-4!-5!/2!-0+0\\=2\,916}$$
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .