<p>A singer is selecting musicans to serve as her back-up band on her latest tour. She must select exactly two guitarists, one bassist, and one drummer. Of the musicians she can choose from to accompany her, five play guitar, two play bass, and four play drums. How many different back-up bands can she create?</p>
<p>(A) 40
(B) 80
(C) 120
(D) 240
(E) 330</p>
<p>She needs two guitarists, and there are 5 to choose from. For her first choice, she has 5 different possibilities, and for her second, she has four.
5 x 4</p>
<p>She needs one bassist, and there are two to choose from. For her third choice, she has two different possibilities.
5 x 4 x 2.</p>
<p>She needs one drummer, and there are four to choose from. For her fourth choice, she has four different possibilities.
5 x 4 x 2 x 4 = 160... am I missing something?</p>
<p>seems correct to me</p>
<p>( 5! / [ 2! (5-2)! ] ) x 2 x 4
= 80
= B</p>
<p>Wait fusion why do you need to do that? It is B</p>
<p>I would do it (5*4)/2 * 2 * 4 = 80, but same thing as Fusion.</p>
<p>ILoveBrown's method doesn't account for duplicates. There are 10 different combinations of guitarists, not 20, because order is unimportant. If the guitarists are Joe, James, Jack, Jared, and Jim, ILoveBrown's method counts Joe/James and James/Joe as two different pairings.</p>
<p>Fusion's way is the technical way to do it, but you just need to divide by two to eliminate duplicates.</p>
<p>Ah, thanks bigmrpig. Good think I took the SAT three years ago when I was sharp!</p>
<p>a classic combination problem .. 5C2<em>2C1</em>4C1 = 10<em>2</em>4</p>
<p>explanation:
5 guitarist but only can choose 2 = 5C2
etc....</p>
<p>Hope my explanation was clear and concise ;)</p>