document.write( "Question 430300: In a certain coin-flipping game, the player wins if he/she gets an even number of heads in five tries. How many ways are there to win this game? (Note: 0 is an even number).
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Algebra.Com's Answer #298828 by richard1234(7193) $\"\"$ $\"About$

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In total, the player can get 0,1,2,3,4, or 5 heads. Let n(x) denote the number of ways to obtain x heads. We can see that n(0) = n(5), n(1) = n(4), and n(2) = n(3) by a symmetry argument. There are 2^5 = 32 total possible outcomes. Since n(0) + n(2) + n(4) = n(1) + n(3) + n(5), and $\"sum%28n%28i%29%2Ci=0%2C5%29+=+32\"$ , then n(0) + n(2) + n(4) = 16. \n" ); document.write( "

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