document.write( "Question 371024: help me solve this problem please. I don't know how to get the answers.\r
\n" ); document.write( "\n" ); document.write( "a) How many different 7-digit numbers can be formed from the digits 0,1,2,2,3,3,3 assuming that a number cannot start with 0?\r
\n" ); document.write( "\n" ); document.write( "ans : 360\r
\n" ); document.write( "\n" ); document.write( "b) how many of these numbers will end in 0?\r
\n" ); document.write( "\n" ); document.write( "ans : 60
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Algebra.Com's Answer #264856 by sudhanshu_kmr(1152)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "2 is 2 times and 3 is 3 times repeated in given digits.\r
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\n" ); document.write( "\n" ); document.write( "a) no. of ways to form 7 digit no. = 7! / ( 2! * 3!) = 420\r
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\n" ); document.write( "\n" ); document.write( " no. of ways to form 7 digit no. having 0 at starting place = 6!/(2! * 3!) = 60\r
\n" ); document.write( "\n" ); document.write( "( here we have to arrange remaining 6 digits only because 0 at fixed position )\r
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\n" ); document.write( "\n" ); document.write( "so, total no. that not start with 0 = 420-60 = 360 \r
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\n" ); document.write( "\n" ); document.write( "b)similarly to last solution\r
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\n" ); document.write( "\n" ); document.write( "no. of ways to form 7 digit no. having 0 at last place = 6!/(2! * 3!)\r
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