Find the number of four digit even numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of those will be even?
Hint: We use the principle of permutations and combinations to find the number of four digit even numbers using digits 1, 2, 3, 4 and 5. Thus, we use the fact that each digit has five possibilities and at the same time keeping in mind that we don’t repeat the digits. Further, to ensure that the number is even we ensure that in the unit’s place, the digits can only be 2 or 4.
Complete step-by-step answer: First, since we have a constraint for the unit's place, we start by counting the combinations for the unit's place. Thus, we have only two possibilities – 2 or 4. Now, for the remaining places, we can start from any digit’s place to calculate the combinations. Suppose, we start with the tenths place, we now have five possibilities (1, 2, 3, 4 and 5) but we have to exclude one possibility since the unit’s place is already filled and no digit can be repeated. Thus, we would have four possibilities. Now, we move on to hundredths place, we would similarly have only 3 possibilities (since, now two places are filled and no digits can be repeated). Similarly, we would now have only two possibilities for the thousandths place. Thus, to calculate the total number of four digit even numbers, we multiply all possibilities to get the answer. Thus, we have, total possibilities-
2 $\times $ 4 $\times $ 3 $\times $ 2 = 48
Hence, there are 48 four digit even numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated.
Note: Generally, for questions regarding permutations and combinations involving calculations of number of digits, we start with the constraint portion of the problem. In this case, we start with the fact that we required the four digit number to be even and then proceed forward. This greatly helps us in simplifying the problem and further prevents from incorrectly counting the total numbers.
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