<p>Here's the question from Section 2 question 8, January 2007:</p>
<p>"Meredith has a red hat, a blue hat , and a white hat. She also has three sweaters- one red, one blue, and one white- and three pairs of jeans- one red, one blue, and one white. Meredith wants to wear a red, white, and blue outfit consisting one one hat, one sweater, and one pair of jeans. How many different possibilities does she have?
a) 3 b)6 c) 9 d) 12 e) 27</p>
<p>Please help out! Is there any method besides making a list that could solve this problem (i.e combinations /permutations) ? And if the list is the method, could someone systematically show how to list them?</p>
<p>Another guestion (though less hard ) ifrom the Jan. 07 test section 9, question 12:</p>
<p>"In the figure above, rectangles PQRS and WXYZ each have perimeter 12 and are inscribed in the circle. How many other rectangles with perimeter 12 can be inscribed in the circle.?"</p>
<p>Basically, the diagram is just a circle with 2 same looking rectangles going through it one vertically, and the other horizontally.</p>
<p>The answer is "more than four "</p>
<p>Would appreciate help for both of these questions. Thanks again.</p>
<p>Meredith has a red hat, a blue hat , and a white hat. She also has three sweaters- one red, one blue, and one white- and three pairs of jeans- one red, one blue, and one white. Meredith wants to wear a red, white, and blue outfit consisting one one hat, one sweater, and one pair of jeans. How many different possibilities does she have?</p>
<p>IF the hat is red, she can wear either (blue jeans, white sweater OR vice versa)
Similary, if the hat is white , she can wear (red jeans, blue sweater OR vice versa)
So with every hat, 2 combos are possible. and there are 3 hats.
So 6 combinations.</p>
<p>"In the figure above, rectangles PQRS and WXYZ each have perimeter 12 and are inscribed in the circle. How many other rectangles with perimeter 12 can be inscribed in the circle.?"</p>
<p>Basically, the diagram is just a circle with 2 same looking rectangles going through it one vertically, and the other horizontally.</p>
<p>Infinite number of rectangles with the same perimeter can be inscribed in the circle... Since the vertical rectangle and the horizontal rectangle are the same , but just at different angles... So u can have infinite such rectangles at varying inclinations.</p>
<p>spidey...u rock,man.....well,i i am not fan of listing,so i use combination....here is how i would like to solve.....forgive my butting in....</p>
<p>let's choose any of hat,sweater or jeans....i choose jeans...now i have 3 choices for jeans--- red,blue,white.....when i pick up a definite color of jeans (say, red) then i have 2 options of choosing the sweater which r either blue or white....so i pick up blue sweater and only white hat is left for me.....so the combination would be 3X2X1=6....so answer is 6</p>
<p>by the way,spidey gave a killer solution for the second question....interjecting anything else ll b stupid</p>
<p>thanks for the help! quick question about/ for the last post: Is that a combination method? or is it just a "multiplication counting principle" method? I don't want to confuse these!</p>