SOLUTION: (a)using the digits 2,3,4,5,6,7 (I) how many three number digit numbers can be formed if repetition is not permitted? (ii) how many of these 3-digit numbers would be odd? (iii)h

Algebra ->  Finance -> SOLUTION: (a)using the digits 2,3,4,5,6,7 (I) how many three number digit numbers can be formed if repetition is not permitted? (ii) how many of these 3-digit numbers would be odd? (iii)h      Log On


   

Question 1118511: (a)using the digits 2,3,4,5,6,7
(I) how many three number digit numbers can be formed if repetition is not permitted?
(ii) how many of these 3-digit numbers would be odd?
(iii)how many of these 3-digit numbers would be divisible by 5?
(c) find the term containing x^9 in the expansion of (x^3 - 2/x)^15 without actually doing the expansion
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
There are 6 ways to choose the first digit, 5 the second and 4 the third. There are 120 ways.
Half or 60 will be odd.
divisible by 5, must end in 5, so 6 ways to choose first and 5 ways to choose the second or 30 ways.
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
this is 455[(x^3)^3-(2/x)^12]=455 x^9-(455*4096)/x^12



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


(aI) 6*5*4 = 120 (6 choices for first digit; 5 for second; 4 for third)

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

(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)

(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:

%28x%5E3%29%5E6%2A%28x%5E-1%29%5E9+=+%28x%5E18%29%28x%5E-9%29+=+x%5E9%29

The term is then