With the word Quicker, how many ways can we arrange the letters if U and I can't be together?

1 Answer

1800 ways

Explanation:

Let's start off by seeing that there are 7 letters. If the letters were all different and we didn't have placement issues (U and I can't be set together), we could arrange them in #7! = 5040# ways. Keep this in mind when working this out - our final answer can't be more than #7!#.

With working with the U, I placement restriction, we can approach it by calculating the number of ways we can place U and I and then multiply by the ways we can place the remaining letters.

U, I Placement

One way we can calculate this is to count the number of ways we can place U and I. They can be in places:

#1, 3; 1, 4; 1, 5; 1, 6; 1, 7#
#2, 4; 2, 5; 2, 6; 2,7#
#3, 5; 3, 6; 3, 7#
#4, 6; 4, 7#
#5, 7#

#= 15# ways the seats work. We have to double this number (we can run through the combos once for U to the left, then again with I to the left), so the grand total is 30 ways.

Placing the remainder

We can now place the remaining letters. There are five of them and if they were all different, we could place them in #5! = 120# ways. But there are two Cs, so that will create duplicates. We divide by the number of ways the Cs can be placed (whenever we have any letter placement, #C_1# could be on the left and #C_2# on the right, or it could be the other way). It turns out that we divide by the factorial of the number of elements that are duplicates, and so we divide by #2!#. This gives us:

#(5!)/(2!)=(5xx4xx3xx2!)/(2!)=60#

Putting it together

We multiply the number of ways we can place the U and I with the ways to place the remaining letters:

#30xx60=1800# ways.