+9 I've been having some trouble...
1 - Four German diplomats, three French diplomats, and a British diplomat are to be seated equally spaced around a circular table. How many different arrangements are possible if the Germans must all sit together (all four seated consecutively) and the French must all sit together (all three consecutively)? (As usual, two arrangements are identical if one is a rotation of the other.)
2 - Ellie wants to color two sides of a regular octagon green, two blue, two red, and two yellow such that each pair of opposite sides of the octagon have the same color. How many different patterns can she form? (Two patterns are considered identical if one can be rotated to form the other.)
3 - In poker, a 5-card hand is called two pair if there are two cards of one rank, two cards of another rank, and a fifth card of a third rank. (For example, QQ668 is two pair.) The order of the cards doesn't matter (so, for example, QQ668 and 68QQ6 are the same). How many 5-card hands are two pair? (Assume a standard 52-card deck with 13 ranks in each of 4 suits.)
4 - I roll a fair 6-sided die five times. How many sequences of rolls will give me three 2's, one 4, and one 6? (A "sequence of rolls" is an ordered list of rolls, like 1, 1, 2, 1, 2.)
5 - In how many ways can friends sit at a dinner table if two of them, Anna and Bob, insist on having exactly two people seated between them? (As usual, two seatings are considered the same if one is a rotation of the other.)
+128408
Here's the first one
Anchor the British Diplomat in any position
The Germans can then be arranged in 4! ways = 24 ways and the French can be seated in 3! ways = 6 ways
And the Germans and French themselves can each be seated in 2! ways = 2 ways on either side of the British diplomat....so the total arrangements = 2! * 3! * 4! = 2 * 6 * 24 = 288 arrangements
+128408
Here's the second one
First, just choose any one of the colors to occupy any two opposite sides
Then, the other 3 colors can be arranged in 3! = 6 ways
So......
6 different patterns are possible
+128408
Here's the third one
From any of the 13 ranks we want to choose any 2 of them
And from each rank we want to choose and 2 of 4 cards
And from the other 44 remaining cards, we want to choose 1 of them
So....the total number of pairs are
C(13,2) * C(4,2)^2 * C(44,1) = 123,552
+128408 Here's the fourth one
We are just looking for all the possible arrangememts of this set (2, 2, 2, 4, 6)
Note that the "6" can occupy any one of 5 positions and once its placement is determined, the "4" can placed in any of the 4 other positions.....
So..... the total number of sequences = 5 * 4 = 20
