Question 1118511
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(aI) 6*5*4 = 120  (6 choices for first digit; 5 for second; 4 for third)<br>
(aII) 3*5*4 = 60  (only 3 choices for last digit, because it must be odd; then 5 for second and 4 for third)<br>
(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)<br>
(c) When expanding {{{x^3-2/x)^15}}} to get a term with x^9, you need to take the x^3 term 6 times and the 2/x term 9 times:<br>
{{{(x^3)^6*(x^-1)^9 = (x^18)(x^-9) = x^9)}}}<br>
The term is then
{{{C(15,6)((x^3)^6)(-2/x)^9 = (5005)(-512)x^9 = -2562560x^9}}}