The total number of ways to organize A, B, C, D, and E is 5! = 120.

This, however, has instances that break the rules by placing A+B together. If we reorganized these as A+B, C, D, and E, we can count all the rule-breaking instances as a 4! = 24.

Subtract the rule-breakers from the total, and you get choice (E) 96.

96 seems to be the right answer but there has to be a better way to solve it than below:

Total possibilities = 5! =120 ways
if A,B are in positions 1&2 = 1x1x3x2= 6 ways
similarly 2&3, 3&4, 4&5 = 6×3= 18 ways
Total of 24 ways when A&B are together. Hence 120-24 =96 ways A&B are not together

But I won’t get it right within the time frame on the test.

The answer is D(72)
Without any restrictions in arrangement you get 5*4*3*2*1=120 configurations
There are 2(4*3*2)=24 configurations in which A and B are next to each other
Therefore the answer is 120-48=72

4x4x3x2 shouldnt work because there are two exceptions. If you picked either A or B for the first number [which is why you are going for number 4 instead of 5] then the next number cannot be the forbidden number [so you are picking from C,D,E]. Then you can pick any number from remaining.

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Is this a correct approach?

The total number of ways to organize A, B, C, D, and E is 5! = 120.

This, however, has instances that break the rules by placing A+B together. If we reorganized these as A+B, C, D, and E, we can count all the rule-breaking instances as a 4! = 24.

Subtract the rule-breakers from the total, and you get choice (E) 96.

OK, I get it. All combinations of A+B, C, D, and E are rule-breakers. B+A, C, D, and E are also rule-breakers. 120-24-24 = 72, choice D.

96 seems to be the right answer but there has to be a better way to solve it than below:

Total possibilities = 5! =120 ways

if A,B are in positions 1&2 = 1x1x3x2= 6 ways

similarly 2&3, 3&4, 4&5 = 6×3= 18 ways

Total of 24 ways when A&B are together. Hence 120-24 =96 ways A&B are not together

But I won’t get it right within the time frame on the test.

The answer is D(72)

Without any restrictions in arrangement you get 5*4*3*2*1=120 configurations

There are 2(4*3*2)=24 configurations in which A and B are next to each other

Therefore the answer is 120-48=72

D. 72

96. because there is one exception. Therefore, I multiplied 4X4X3X2X1=96.

4x4x3x2 shouldnt work because there are two exceptions. If you picked either A or B for the first number [which is why you are going for number 4 instead of 5] then the next number cannot be the forbidden number [so you are picking from C,D,E]. Then you can pick any number from remaining.

So 4 x 3 x 3 x 2 = 72

i think the correct answer is E.96

D?

answer is A-24