document.write( "Question 1082187: 1.FIND how many 5 digits odd number greater than 40000 that can be obtained using the digit 1,3,5,6,7,8.
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\n" ); document.write( "Find possible arrangements if NO TWO women are standing next to each other. \n" ); document.write( "
Algebra.Com's Answer #696351 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
1.FIND how many 5 digits odd number greater than 40000
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document.write( "Since the 5-digit numbers must be odd, their last digit is 1, 3, 5 or 7\r\n" );
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document.write( "Case 1.  The last digit is 1 or 3\r\n" );
document.write( "Then the first digit is 5, 6, 7 or 8 in order to be greater than 40000.\r\n" );
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document.write( "We can choose the last digit either of 2 ways, as 1 or 3.\r\n" );
document.write( "We can then choose the first digit any of 4 ways. (as 5, 6, 7 or 8.)\r\n" );
document.write( "We choose the second digit as any of the remaining 4 digits.\r\n" );
document.write( "We choose the third digit as any of the remaining 3 digits.\r\n" );
document.write( "We choose the fourth digit as any of the remaining 2 digits.\r\n" );
document.write( "Answer for case 1:  2×4×4×3×2 = 192 ways\r\n" );
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document.write( "Case 2.  The last digit is 5 or 7\r\n" );
document.write( "We can choose the last digit either of 2 ways, as 5 or 7.\r\n" );
document.write( "We can then choose the first digit any of 3 ways. \r\n" );
document.write( "       (as 6, 8, or the unchosen one of 5,7.)\r\n" );
document.write( "We choose the second digit as any of the remaining 4 digits.\r\n" );
document.write( "We choose the third digit as any of the remaining 3 digits.\r\n" );
document.write( "We choose the fourth digit as any of the remaining 2 digits.\r\n" );
document.write( "Answer for case 2:  2×3×4×3×2 = 144\r\n" );
document.write( "Total for both cases:  192+144 = 336 ways.\r\n" );
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b) repetition allowed
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document.write( "Since the 5-digit numbers must be odd, their last digit is 1, 3, 5 or 7\r\n" );
document.write( "We can choose the last digit any of 4 ways, as 1, 3, 5 or 7.\r\n" );
document.write( "We can then choose the first digit any of 4 ways. (as 5, 6, 7 or 8.)\r\n" );
document.write( "We choose the second digit as any of the 6 digits.\r\n" );
document.write( "We choose the third digit as any of the 6 digits.\r\n" );
document.write( "We choose the fourth digit as any of the 6 digits.\r\n" );
document.write( "4×4×6×6×6 = 3456 ways\r\n" );
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\n" ); document.write( "6 men and 3 women are standing in a supermarket queue.
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document.write( "First we arrange the 6 men in a line in 6! = 720 ways.\r\n" );
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document.write( "Think of the row of men like this with spaces between them,\r\n" );
document.write( "1 space immediately before the leftmost man and 1 space \r\n" );
document.write( "immediately after the rightmost man, like this:\r\n" );
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document.write( "_M_M_M_M_M_M_\r\n" );
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document.write( "The 7 \"_\"'s are potential places where the three women can\r\n" );
document.write( "be inserted among the men.  That's 7P3 = 7×6×5 = 210 ways to \r\n" );
document.write( "insert the 3 women in three of those blanks so that no two \r\n" );
document.write( "women are next to each other.\r\n" );
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document.write( "Answer: 6!×(7P3) = 720×210 = 151200 ways.\r\n" );
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document.write( "Edwin
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