Imaging you are – like me 🙂 – staying for 4 nights at a lovely little Italian hotel, just in front of the awesome Garda Lake, in the north of Italy… A dream come true!

It is really hot… 37°C at least!… It is really summer!…

Since the hotel has an excellent restaurant, you decide to eat there once daily, each of the 3 days in between check-in and check-out. And since it is so hot, you choose only 2 cold starters each day.

- Quiz: If in the menu there are 7 starters you really would like to try, how many ways to order your 3 dinners of 2 starters do you have?

Well, 3 dinners, of 2 starters each, among 7 different plates…

Can you guess??…

In fact, we could solve this just like we do in the Cocktail Quiz! But, *interesting* – and this is a characteristic of math 😉 – if we change a bit the way we see the situation, we can find a maybe different answer to a *different* question: Actually, what we do each time is choosing a “pair” of starters! That is what we really do when we order each day. But with food, we usually care about the order! This is a new math-scenario: The question now is “In how many ways can you order 3 pairs of starters, taking account of the order within the pairs?:

First you chose 2 from the 7. Then you choose another 2 from the resting 5 and then you choose 2 from the resting 3. This gives 7! / 2! x (7 – 2)! times 5! / 2! x (5 – 2)! times 3! / 2! x (3 – 2)! which is

1x2x3x4x5x6x7 / 2 x 1x2x3x4x5 x 1x2x3x4x5 / 2 x 1x2x3 x 1x2x3 / 2 x 1 = 6×7/2 x 4×5/2 x 3 = 21 x 10 x 3 = 630.

And now you should multiply by… 2 x 2 x 2 = 8, the different orders in each of the 3 pairs. This gives a total of

680 x 8 = 5040

Note: This is just the same as considering ALL the possible orders of 6 starters among 7.

Just 4 Fun…