document.write( "Question 1204522: How many odd numbers greater than 70,000 can be formed using the digits 0,1,4,7,8,9.
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document.write( "(a) without repetition.
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document.write( "(b) if repetitions are allowed. \n" );
document.write( "
Algebra.Com's Answer #840814 by greenestamps(13203) You can put this solution on YOUR website! \n" ); document.write( "(a) without repetition.... \n" ); document.write( "(1) 5-digit numbers: \n" ); document.write( "7xxx1 \n" ); document.write( "7xxx9 \n" ); document.write( "8xxx1 \n" ); document.write( "8xxx7 \n" ); document.write( "8xxx9 \n" ); document.write( "9xxx1 \n" ); document.write( "9xxx7 \n" ); document.write( "For each of those 7 cases, the number of different numbers is 4*3*2 = 24. \n" ); document.write( "That's 7*24 = 168 5-digit numbers \n" ); document.write( "(2) 6-digit numbers: \n" ); document.write( "1xxxx7 \n" ); document.write( "1xxxx9 \n" ); document.write( "4xxxx1 \n" ); document.write( "4xxxx7 \n" ); document.write( "4xxxx9 \n" ); document.write( "7xxxx1 \n" ); document.write( "7xxxx9 \n" ); document.write( "8xxxx1 \n" ); document.write( "8xxxx7 \n" ); document.write( "8xxxx9 \n" ); document.write( "9xxxx1 \n" ); document.write( "9xxxx7 \n" ); document.write( "For each of those 12 case the number of different numbers is 4*3*2*1 = 24. \n" ); document.write( "That's 288 6-digit numbers. \n" ); document.write( "ANSWER (without repetition): 168+288 = 456 \n" ); document.write( "(b) with repetition.... \n" ); document.write( "The patterns for the case with repetition allowed are the same; but now for each case the number of different numbers is 6*6*6*6 = 1296. \n" ); document.write( "That makes the total number of different 5- or 6-digit numbers \n" ); document.write( "(7+12)(1296) = 24624 \n" ); document.write( "However, for the case where repetition is allowed, there are in fact an infinite number of odd numbers greater than 70000, because there is no stated restriction on the number of digits. \n" ); document.write( " \n" ); document.write( " |