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# Can someone help me with these?

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1577
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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.)

Jun 12, 2017

#1
+98124
+2

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

Jun 12, 2017
#2
+98124
+3

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

Jun 12, 2017
#3
+98124
+2

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

Jun 12, 2017
#4
+98124
+1

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

Jun 12, 2017