How many ”words” can be formed by rearranging INQUISITIVE so that U does not immediately follow Q?

Question

My attempt:

Step 1: Place the Q: $11$ ways

Step 2: Place the U accordingly: $9$ ways

Step 3: Place the remaining $9$ letters: $9!$ ways

So there are $11\times 9 \times 9!$ arrangements.

Can someone please check.

Answer 1

There are two small mistakes in your calculation:

• Placing Q at the end gives 10 possibilities to place U.

There are $10$ ways of placing Q with $9$ ways of placing U and $1$ way of placing Q with $10$ ways of placing U: $\boxed{10 \cdot 9 + 1 \cdot 10}$

• The permutations of the 4 I's need to be canceled out: $4!$

So, you get $$\frac{(10 \cdot 9 + 1 \cdot 10)\cdot 9!}{4!}$$

Another approach could be:

• All possible arrangements: $\frac{11!}{4!}$
• Arrangements with QU together: $\frac{10!}{4!}$

Subtraction gives the arrangements where U does not directly follow Q: $$\frac{11!}{4!} - \frac{10!}{4!} = \frac{11!-10!}{4!}$$

Answer 2

• All possible arrangements: $\frac{11!}{4!}$
• Arrangements with QU appear=$\frac{\text{ Arrangements where Q and U appear together}}{2}=\frac{10!}{2.4!}$

Subtraction gives the arrangements where U does not directly follow Q: $$\frac{11!}{4!} - \frac{10!}{2.4!} = \frac{11!-\frac{10!}{2}}{4!}$$