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 #733883 by greenestamps(13200)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "(aI) 6*5*4 = 120 (6 choices for first digit; 5 for second; 4 for third)

\n" ); document.write( "(aII) 3*5*4 = 60 (only 3 choices for last digit, because it must be odd; then 5 for second and 4 for third)

\n" ); document.write( "(aIII) 1*5*4 = 20 (only 1 choice for last digit, because the number has to be divisible by 5; then 5 for second and 4 for third)

\n" ); document.write( "(c) When expanding \"x%5E3-2%2Fx%29%5E15\" to get a term with x^9, you need to take the x^3 term 6 times and the 2/x term 9 times:

\n" ); document.write( "\"%28x%5E3%29%5E6%2A%28x%5E-1%29%5E9+=+%28x%5E18%29%28x%5E-9%29+=+x%5E9%29\"

\n" ); document.write( "The term is then
\n" ); document.write( "
\n" ); document.write( "
\n" );