document.write( "Question 1118511: (a)using the digits 2,3,4,5,6,7
\n" ); document.write( "(I) how many three number digit numbers can be formed if repetition is not permitted?
\n" ); document.write( "(ii) how many of these 3-digit numbers would be odd?
\n" ); document.write( "(iii)how many of these 3-digit numbers would be divisible by 5?
\n" ); document.write( "(c) find the term containing x^9 in the expansion of (x^3 - 2/x)^15 without actually doing the expansion \n" ); document.write( "
Algebra.Com's Answer #733864 by Boreal(15235)\"\" \"About 
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There are 6 ways to choose the first digit, 5 the second and 4 the third. There are 120 ways.\r
\n" ); document.write( "\n" ); document.write( "Half or 60 will be odd.\r
\n" ); document.write( "\n" ); document.write( "divisible by 5, must end in 5, so 6 ways to choose first and 5 ways to choose the second or 30 ways.\r
\n" ); document.write( "\n" ); document.write( "The third term will be the cube of x^3 and will yield x^9. The third term has coefficient of 15C3=15*14*13*12!/12!*3*2*1=455
\n" ); document.write( "this is 455[(x^3)^3-(2/x)^12]=455 x^9-(455*4096)/x^12\r
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