How many 4-digit even number can be formed by using the digits 5, 7 9 and 2 only once
|
Without considering different cases. Show
This is another way to get the same answer as already answerd above. Here we start by finding the total amount of the four-digit numbers with distinct digits, then finding the amount of odd digits, filling the same criteria, and last, we subtract the odd numbers from the total, to find the even numbers filling the criteria. We have totally seven digits. We know that the digit zero cannot be placed at the position representing the thousand position. That leaves us with six digits to choose from for this position and that can be made in $P(6,1) = \frac{6!}{(6-1)!} = \frac{6!}{5!}=6$, different ways. Now, we have three positions left to fill and six digits to choose from, including the digit zero, which can be placed anywhere in the remaining positions. This choice can be done in $P(6,3) = \frac{6!}{(6-3)!} = \frac{6!}{3!}=6 \cdot5 \cdot4$, different ways. Finally, with distinct digits, there is $6 \cdot6 \cdot5 \cdot4 =720$ four-digit numbers to be constructed. We know that the amount of odd numbers plus the amount of even numbers equal the total amount of the 720 four-digit numbers. The odd numbers are 1, 3 and 5. The question to be asked is how many of the 720 are odd? The digit to fill the unit position can only be chosen from the digits 1, 3 or 5, and this choice can be made in $P(3,1)=\frac{3!}{(3-1)!}=\frac{3!}{2!}=3$, different ways. To choose the digit filling the thousand position, we have five valid digits to choose from, since the zero digit is not valid for this position. The choice can be made in $P(5,1)=\frac{5!}{(5-1)!}=\frac{5!}{4!}=5$, different ways. Now we are left with two positions to fill, the hundred and tenth position, and five digits to choose from, now including the zero digit. The choice for these two positions can be made in $P(5,2)=\frac{5!}{(5-2)!}=\frac{5!}{3!}=5 \cdot4=20$, different ways. The total number of odd four-digit numbers, with distinct digits are $5 \cdot5 \cdot4 \cdot3 =300$. Now, we can answer the question how many of these 720, four-digit numbers, are even, by the subtraction $720-300=420$. The even numbers are 420. In mathematics, permutation is known as the process of arranging a set in which all the members of a set are arranged into some series or order. The process of permuting is known as the rearranging of its components if the set is already arranged. Permutations take place, in more or less important ways, in almost every area of mathematics. They frequently appear when different commands on certain finite sets are considered. What is a Combination? A combination is an act of choosing items from a group, such that (not like permutation) the order of choice does not matter. In smaller cases, it is possible to count the number of combinations. Combination refers to the union of n things taken k at a time without repetition. In combination, you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used. Permutation Formula In permutation r things are selected from a set of n things without any replacement. In this order of selection matter.
Combination Formula In combination r things are selected from a set of n things and where the order of selection does not matter.
How many 3-digit even numbers can be formed by using the digits 1,2,3,4, and 5?Solution:
Similar QuestionsQuestion 1: How many 3 digit odd numbers can be formed by using the digits 1,2,3,4 and 5? Solution:
Question 2: How many 4 digit even numbers can be formed by using the digits 1,2,3,4 and 5? Solution:
How many 4 digit even number can be formed by using the digits 5 7 9 and 2 only once immersive reader?∴ The required result will be 60.
How many 4 digits numbers can be formed using digits 1 3 4 5 7 9 repetition not allowed?Hence, there are 720 numbers can be formed.
How many 4 digit even numbers can be formed using the digits?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.
How many 4 digits even numbers can be formed from the digits 5 6 7 8 9 if no digits repeated?So, total 360 four-digit numbers can be formed from the given digits where a digit is used at most once.
|
