alfonspm wrote:
How many even numbers, larger than 400, with 3 different digits, can be formed using digits 1, 2, 3, 4, 5, 6, 7, 8?
A. 102
B. 620
C. 640
D. 2048
E. 2520
I am struggling to understand why 102 is the correct answer as my approach is; 5 options for the first digit, 7 for the second and from there it is already wrong because there is no integer to multiply 5*7 that result in 102. Thank you in advance
P.D.: I am sorry for any possible mistake as I am new in the forum. I could not find the right topic for the problem because I have it as a counting methode. The source is from an online platform provided by my Academy.
1. If the hundreds digit is even (so 4, 6, or 8):Number of options for the hundreds digit: 3;
Number of options for the units digit: 3 (any from 2, 4, 6, or 8 but one used for the hundreds);
Number of options for the tens digit: 6.
Total = 3*3*6 = 54.
2. If the hundreds digit is odd (so 5, or 7):Number of options for the hundreds digit: 2;
Number of options for the units digit: 4 (any from 2, 4, 6, or 8);
Number of options for the tens digit: 6.
Total = 2*4*6 = 48.
GRAND TOTAL = 54 + 48 = 102.
Answer: A.
alfonspm wrote:
P.D.: I am sorry for any possible mistake as I am new in the forum. I could not find the right topic for the problem because I have it as a counting methode. The source is from an online platform provided by my Academy.
The Category for such kind of questions in Combinations.
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