document.write( "Question 76806: How many different 6-player starting squads can be formed from a volleyball team of 11 players? \n" ); document.write( "
Algebra.Com's Answer #55055 by Edwin McCravy(20056) $\"\"$ $\"About$
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document.write( "How many different 6-player starting squads can be formed from a volleyball\r\n" );
document.write( "team of 11 players?\r\n" );
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document.write( "Let's say the 11 players are A,B,C,D,E,F,G,H,I,J,K\r\n" );
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document.write( "First we figure how many ORDERED groups (permutations) of 5 players there are,\r\n" );
document.write( "where one such ORDERED group would be. say,\r\n" );
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document.write( "G,H,B,A,J,F\r\n" );
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document.write( "We could have chosen the first one (where the G is) any of 11 ways.\r\n" );
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document.write( "Then we could have chosen the second one (where the H is) any of the remaining\r\n" );
document.write( "10 ways. That's (11)(10) ways.\r\n" );
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document.write( "Then we could have chosen the third one (where the B is) any of the remaining 9\r\n" );
document.write( "ways. That's (11)(10)(9) ways.\r\n" );
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document.write( "Then we could have chosen the fourth one (where the A is) any of the remaining\r\n" );
document.write( "8 ways. That's (11)(10)(9)(8) ways.\r\n" );
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document.write( "Then we could have chosen the fifth one (where the J is) any of the remaining 7\r\n" );
document.write( "ways. That's (11)(10)(9)(8)(7) ways.\r\n" );
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document.write( "Then we could have chosen the sixth one (where the F is) any of the remaining 6\r\n" );
document.write( "ways. That's (11)(10)(9)(8)(7)(6) ways. \r\n" );
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document.write( "That would be the answer if they had to be in a certain order.\r\n" );
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document.write( "However we know that they don't have to be in any certain order. For example,\r\n" );
document.write( "the ordered group G,H,B,A,J,F is not really a different group from the ordered\r\n" );
document.write( "groups A,G,F,J,H,B or J,B,A,G,F,H or any other arrangement of those 6 players. \r\n" );
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document.write( "So therefore\r\n" );
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document.write( "(11)(10)(9)(8)(7)(6) is entirely too large a number. Let's look at the sample\r\n" );
document.write( "ordered group G,H,B,A,J,F to see how many times it is counted among the\r\n" );
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document.write( "(11)(10)(9)(8)(7)(6), so we can divide by that number.  \r\n" );
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document.write( "We could have chosen the first one (where the G is) any of 6 ways.\r\n" );
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document.write( "Then we could have chosen the second one (where the H is) any of the remaining\r\n" );
document.write( "5 ways. That's (6)(5) ways.\r\n" );
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document.write( "Then we could have chosen the third one (where the B is) any of the remaining 4\r\n" );
document.write( "ways. That's (6)(5)(4) ways.\r\n" );
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document.write( "Then we could have chosen the fourth one (where the A is) any of the remaining\r\n" );
document.write( "3 ways. That's (6)(5)(4)(3) ways.\r\n" );
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document.write( "Then we could have chosen the fifth one (where the J is) any of the remaining 2\r\n" );
document.write( "ways. That's (6)(5)(3)(2)(1) ways.\r\n" );
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document.write( "Then we could have chosen the sixth one (where the F is) ONLY the remaining 1\r\n" );
document.write( "way. That's (6)(5)(4)(3)(2)(1) ways.\r\n" );
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document.write( "Since every group of 6 is counted among the (11)(10)(9)(8)(7)(6) ORDERED groups\r\n" );
document.write( "(6)(5)(4)(3)(2)(1) times, the desired number is \r\n" );
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document.write( "    \r\n" );
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document.write( "Then you can have a good time canceling the factors of the denominator\r\n" );
document.write( "into the factors of the numerator and end up with 462.\r\n" );
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document.write( "The answer could have been done as C(11,6) and the formula\r\n" );
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document.write( " used as\r\n" );
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document.write( " =\r\n" );
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document.write( "Edwin
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