Pietrus95 wrote:
Can someone help with this problem??
I am thinking that both cases can lead to understanding how many sets are possible.
For instance, for rule 1 i can have the following set:
1324 or even 3214 etc. leading to 4!. I am also capable of finding the total number of combinations which can have the sum equal to 10. Same for the second rule.
What is the answer?
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N is a four-digit number. N = ABCD where A, B, C, D are distinct single-digit whole numbers. How many possible values of N exist for any fixed set of four values of A, B, C, D?
(1) A + B + C + D = 10if we assume, A = 1, B = 2, C = 3, and D = 4, then (1) is satisfied, and we get 4X3X2X1 = 24 four-digit numbers.
Now, we assume,A = 0, B = 1, C = 2, and D = 7, then (1) is satisfied, and we get 3x3x2x1 = 18 four -digit numbers
so, (1) alone is not sufficient.
(2) A + B + C - D = 8we can use the above approach to say (2) alone is not sufficient.
Now, (1) + (2), we get
A+B+C = 9, and D = 1
2+3+4 = 9, and D = 1, so we get 4x3x2x1 = 24 four-digit numbers
0+3+6 = 9, and D = 1, so we get 3x3x2x1 = 18 four-digit numbers,
so, we don't get a unique number of four-digit numbers, Answer is E.