Question 1190087: A 4-digit number is selected from the numbers {1,2,3,4,5,6} where the digits are selected without replacement.
How many 4-digit numbers can be chosen that are even and greater than 4000?
Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! There are 3 ways out of 6 to get the first digit 4,5, or 6
Any digit can be the second or the third.
The fourth has to be 2, 4, or 6, again a 3/6 probability
The overall probability is (1/2)^2=1/4
Answer by ikleyn(52777)  (Show Source):
You can put this solution on YOUR website! .
A 4-digit number is selected from the numbers {1,2,3,4,5,6} where the digits are selected without replacement.
How many 4-digit numbers can be chosen that are even and greater than 4000?
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The traditional and the standard formulation and the meaning of the problem is that the digits are used without repetition . . .
In this problem, there are two restrictions:
(a) the first (mostleft, thousands) digit must be 4, or 5, or 6;
(b) the last digit (ones digit) must be 2, or 4, or 6 in order for the number be even.
These restrictions are not independent; therefore, an accurate analysis is required.
Case 1. Let the last digit be 2.
Then the leading digit can be 4 or 5 or 6, giving 3 options;
the second digit can be any of remaining 6-2 = 4 digits, giving 4 options;
the third digit can be any of remaining 6-3 = 3 digits, giving 3 options.
Thus in this case we have 3 * 4 * 3 = 36 possible 4-digit numbers.
Case 2. Let the last digit be 4.
Then the leading digit can be 5 or 6, giving 2 options;
the second digit can be any of remaining 6-2 = 4 digits, giving 4 options;
the third digit can be any of remaining 6-3 = 3 digits, giving 3 options.
Thus in this case we have 2 * 4 * 3 = 24 possible 4-digit numbers.
Case 3. Let the last digit be 6. (In this case the analysis is similar to case 2).
Then the leading digit can be 4 or 5, giving 2 options;
the second digit can be any of remaining 6-2 = 4 digits, giving 4 options;
the third digit can be any of remaining 6-3 = 3 digits, giving 3 options.
Thus in this case we have 2 * 4 * 3 = 24 possible 4-digit numbers.
Thus the total of cases 1, 2 and 3 is 24 + 36 + 24 = 84 possible 4-digit numbers.
ANSWER. There are 84 4-digit numbers under imposed conditions.
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