seven guests go to a restaurant and there are 4 different selections from the menu that they can choose from. the waiter takes the order from the seven guests and when the waiter goes back to the kitchen he forgets what the guests had ordered. The waiter tells the chef to make 7 meals. what are the chances they got the order right? Assume the waiter brings the meals out and hands them out all at once.

Lets call the 7 guests a,b,c,d,e,f,g

abc | de | f | g

above you can see that the 7 guests are seperated into 4 sections (the number of choices from the menu)

there are 7 guests and 3 seperators; giving 10 objects.

now it should be easy to show the total number of outcomes

This post was edited by J0nn on Oct 23 2014 09:52pm

The waiter must only choose the correct number of each dish to order since he brings them all out at once to the guests.

You need to know how many ways there are to separate 7 guests into 4 groups. Having noone in a group IS AN OPTION.

http://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29

Using theorem 2, which separates 7 people into 4 bins (that can be zero), we come up with: n = 7, k = 4, (n+k-1) choose (n), or 10 choose 7, or 120.

The answer is 1/120.