Bunuel wrote:

How many different arrangements of A, B, C, D, and E are possible where A is not adjacent to B and D is not adjacent to E?

A. 96

B. 48

C. 32

D. 30

E. 1

This is a hard question.

We'll calculate it by calculating all the possible arrangements, subtracting the 'bad' arrangements and correcting for double-counting.

This is similar to calculating a difficult probability by calculating 1 minus the probability and is a Precise approach.

There are 5! = 120 total options to arrange the 5 letters.

The 'bad' arrangements are those that have an A and a B adjacent or a D and an E adjacent.

To calculate this we'll first 'stick' A and B together to form 'one letter'.

Then we have 4 new letters - 'AB', C, D, E.

There are 4! = 24 ways to arrange these letters

Similarly, each of the combinations of

'BA', C, D, E

A, B, C, 'DE'

A, B, C, 'ED'

has 24 options giving a total of 24*4 = 96 'bad' options.

So far we've counted 120 - 96 = 24 options.

But - we've doublecounted! We've removed arrangements with both A,B adjacent and D,E adjacent twice!

We'll add these in similarly to what we did above:

If 'AB' and 'DE' are both together then we have 3 'letters' we can rearrange:

'AB', C, 'DE' : 3! = 6 options.

'AB', C, 'ED' : 3! = 6 options.

'BA', C, 'DE' : 3! = 6 options.

'BA', C, 'ED' : 3! = 6 options.

So we need to add back 6*4 = 24 giving a final total of 24 + 24 = 48 options.

(B) is our answer.

** It is also possible to brute force list everything but with 5 letters and all the constraints it is probably more dangerous than useful.

_________________

David

Senior tutor at examPAL

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