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Bunuel
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Total cities = 6

Select 4 out of 6: \(^6{C_4} = ^6{C_2} = 15\)

There is 1 way to select A and B together and remaining 2 cities will be selected from 4: \(^4{C_2} = 6 = 1 * 6 = 6\)

A and B not together: 15 - 6 = 9

Answer D
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Bunuel
A fast-food company plans to build four new restaurants. If there are six sites A, B, C, D, E and F, that satisfy the company’s criteria for location of the new restaurants, in how many different ways can the company select four sites if the order of selection does not matter, given that both the sites A and B cannot be selected simultaneously?

(A) 4
(B) 5
(C) 6
(D) 9
(E) 15

The total ways of selecting 4 cities = 15 or 6c4

The total ways of chosing with A and B chosen = 1*4c2=6

Therefore total way of chosing without picking Aand B = 15-6 =9

Therefore IMO D
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Bunuel
A fast-food company plans to build four new restaurants. If there are six sites A, B, C, D, E and F, that satisfy the company’s criteria for location of the new restaurants, in how many different ways can the company select four sites if the order of selection does not matter, given that both the sites A and B cannot be selected simultaneously?

(A) 4
(B) 5
(C) 6
(D) 9
(E) 15

IMO D

Choose 4 sites from the given 6: A,B,C,D,E,F
A & B cannot be together in a chosen set

Option 1: A _ _ _ -> 4C3 ways to fill the remaining spots = 4 ways
Option 2: B _ _ _ -> 4C3 ways to fill the remaining spots = 4 ways
Option 3 (No A & B) : _ _ _ _ -> Only have 1 way (C,D,E,F in any order)

So, 4+4+1 = 9 ways
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My 2 cents on this:

Note down every aspect. 6 locations/sites and 4 restaurants. Total? 6c4= 15 (Therefore answer cannot be E ever)

Then, we are given the condition, A&B cannot be together. So if we equate Total - A&B together = A&B Not together.

A&B together is straightforward. Consider AB is a block that cannot be separated. AB | CDEF are the blocks. AB will give us 2 restaurants. 2c2. Then CDEF will give us 4c2 = 6

So 15-6 = 9 is the answer.

Tricky, but I'm trying to implement this more and more.
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