How many 4 digit even numbers can be formed from the digits 1, 2, 3, 4, 5,6

Answer

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.

Without considering different cases.

Nội dung chính Show

• How many 4 digit pins can be formed in using the digits 1 3 5 and 9 if repetition is not allowed?
• How many four digit even numbers can be formed using the digits 2 3 1 5 7 9 repetition of digits is not allowed?
• How many 4 digit numbers can be formed from the numbers 1 3 5 7 8 and 9 which are divisible by 2 and none of the digits are repeated?
• How many different 4 digit even numbers can be formed from the digits?

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.

Answer

Verified

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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Think of "P" for "permutation" and "P" for "position".

1,3,5,6,8, and 9

There are 6 things: {1,3,5,6,8,9}.
There are 4 Positions or Places to Put them: 
{thousands, hundreds, tens, ones}

That's 6 things to Position in 4 places.

6 POSITION 4 = 6P4 = 6*5*4*3 = 360 such numbers:

[Start with 6, multiply by 1 less each time until 
there are 4 factors. Multiply them all together.]

Explanation:
There are 6 Positions to Put the 1st number. 
That leaves 5 Positions to Put the 2nd number.
That leaves 4 Positions to Put the 3rd number.
That leaves 3 Positions to Put the 4th number.

Here are all 360 4-digit numbers, 36 rows of 10 each,
computer-generated.

1256,1258,1259,1265,1268,1269,1285,1286,1289,1295,
1296,1298,1526,1528,1529,1562,1568,1569,1582,1586,
1589,1592,1596,1598,1625,1628,1629,1652,1658,1659,
1682,1685,1689,1692,1695,1698,1825,1826,1829,1852,
1856,1859,1862,1865,1869,1892,1895,1896,1925,1926,
1928,1952,1956,1958,1962,1965,1968,1982,1985,1986,
2156,2158,2159,2165,2168,2169,2185,2186,2189,2195,
2196,2198,2516,2518,2519,2561,2568,2569,2581,2586,
2589,2591,2596,2598,2615,2618,2619,2651,2658,2659,
2681,2685,2689,2691,2695,2698,2815,2816,2819,2851,
2856,2859,2861,2865,2869,2891,2895,2896,2915,2916,
2918,2951,2956,2958,2961,2965,2968,2981,2985,2986,
5126,5128,5129,5162,5168,5169,5182,5186,5189,5192,
5196,5198,5216,5218,5219,5261,5268,5269,5281,5286,
5289,5291,5296,5298,5612,5618,5619,5621,5628,5629,
5681,5682,5689,5691,5692,5698,5812,5816,5819,5821,
5826,5829,5861,5862,5869,5891,5892,5896,5912,5916,
5918,5921,5926,5928,5961,5962,5968,5981,5982,5986,
6125,6128,6129,6152,6158,6159,6182,6185,6189,6192,
6195,6198,6215,6218,6219,6251,6258,6259,6281,6285,
6289,6291,6295,6298,6512,6518,6519,6521,6528,6529,
6581,6582,6589,6591,6592,6598,6812,6815,6819,6821,
6825,6829,6851,6852,6859,6891,6892,6895,6912,6915,
6918,6921,6925,6928,6951,6952,6958,6981,6982,6985,
8125,8126,8129,8152,8156,8159,8162,8165,8169,8192,
8195,8196,8215,8216,8219,8251,8256,8259,8261,8265,
8269,8291,8295,8296,8512,8516,8519,8521,8526,8529,
8561,8562,8569,8591,8592,8596,8612,8615,8619,8621,
8625,8629,8651,8652,8659,8691,8692,8695,8912,8915,
8916,8921,8925,8926,8951,8952,8956,8961,8962,8965,
9125,9126,9128,9152,9156,9158,9162,9165,9168,9182,
9185,9186,9215,9216,9218,9251,9256,9258,9261,9265,
9268,9281,9285,9286,9512,9516,9518,9521,9526,9528,
9561,9562,9568,9581,9582,9586,9612,9615,9618,9621,
9625,9628,9651,9652,9658,9681,9682,9685,9812,9815,
9816,9821,9825,9826,9851,9852,9856,9861,9862,9865.

Edwin

How many 4 digit pins can be formed in using the digits 1 3 5 and 9 if repetition is not allowed?

Hence, the number of required numbers =(4+12+24+24)=64.

How many four digit even numbers can be formed using the digits 2 3 1 5 7 9 repetition of digits is not allowed?

∴ The required result will be 60.

How many 4 digit numbers can be formed from the numbers 1 3 5 7 8 and 9 which are divisible by 2 and none of the digits are repeated?

How many four digit numbers can be formed from the numbers 1,3,5,7,8 and 9 which are divisible by 2 and none of the digits are repeated? 1. 60.

How many different 4 digit even numbers can be formed from the digits?

So, required number of ways in which four digit even numbers can be formed from the given digits is 2×4×3×2=48.

How many 4 digit even numbers can be formed using the digits 1,2 3 4 and 5 if your not allowed to repeat any digits?

How many 4 digit even numbers are possible by using the digits 1,2 3 4 if none of the digits is repeated?

How many 4 digit numbers can be formed with digits 1,2 3 and 4 and with distinct digits?

What is the Probability of making even 4 digits using 1,2 3 4?