Sum of all 4 digit numbers containing 2, 4, 6, 8 without repetition

Complete step-by-step answer:
The first thing we must find out is the total number of possible arrangements of the 4 digits of the given 5, as the number of ways in which 4 out of the given 5 digits can be arranged is the number of 4 digit numbers formed. Therefore, the number of 4 digit numbers formed from the given 5 digits is $ ^{5}{{P}_{4}} $.
So, we are sure that each of the places of the numbers, i.e., ones place, tens place, hundreds place and thousands place have \[\dfrac{^{5}{{P}_{4}}}{5}\] number of times a digit appearing.
So, the sum of ones digit of all the numbers is: $ \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right) $
Similarly, tens digit will have the same sum, but we need to multiply the result by 10, as the place value of tens place is ten. So, the value of tens place addition is: $ \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 10 $
Similarly, the 100s place value is: $ \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 100 $ and 1000s place value is $ \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 1000. $
Now we will add the place values of all the places to get the answer.
Therefore, the sum of all the numbers satisfying the above condition is:
\[\begin{align}
  & \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)+\dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 10+ \\
 & \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 100+\dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 1000 \\
\end{align}\]
 $ =\dfrac{^{5}{{P}_{4}}}{5}\times 15{{+}^{5}}{{P}_{4}}\times 15\times 10+\dfrac{^{5}{{P}_{4}}}{5}\times 15\times 100+\dfrac{^{5}{{P}_{4}}}{5}\times 15\times 1000 $
 $ =\dfrac{^{5}{{P}_{4}}}{5}\times 15\left( 1+10+100+1000 \right) $
 $ =\dfrac{^{5}{{P}_{4}}}{5}\times 15\times 1111 $
Now we know that $ ^{5}{{P}_{4}}=\dfrac{5!}{1!}=120 $ .
\[\dfrac{^{5}{{P}_{4}}}{5}\times 15\times 1111=\dfrac{120}{5}\times 15\times 1111=399960\]
Therefore, the answer to the above question is 399960.

Note: Remember if one of the digits is zero, then all the digits will not have the same cases of appearing in a given place, actually zero cannot appear at the 1000s place, as if it appears at 1000s place, the number will be a 3 digit number. So, in this case you have to fix zero at every place and get the number of times zero is appearing at each place and divide the left out cases among other digits.

Solution:

There are 4!=24 numbers. Each digit occurring 3!=6 times, in the unit's, ten's, hundred’s and thousand's places. We note that 6(2+4+6+8)=120. Thus in the over all sum there will be 120 units, 120 tens, 120 hundreds and 120 thousands.∴ The required sum =120(1+10+102+103)=120×1111=133320.

Find the sum of all the number formed by 2,4,6, and 8 without repetition.Number may be of any digit like 2, 24, 684, 4862.

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My Approach:

single digit no formed = 2,4,6,8

sum= 2+4+6+8= 20

two digit= 24,26,28,42,46,48,62,64,68,82,84,86

sum= 660

three digit no=246+264+426+462+624+642=2664

268+286+628+682+826+862=3552

248+284+428+482+824+842=3108

468+486+648+684+846+864=3996

sum of all 3 digit nos =13310

Similarly for all 4 digit numbers.

1. Tardigrade
2. Question
3. Mathematics
4. The sum of all 4 digit numbers that can be formed by using the digits 2, 4, 6, 8 (repetition of digits not allowed) is

A

B

C

D

Solution:

There are numbers. Each digit occurring times, in the unit's, ten's, hundred’s and thousand's places. We note that . Thus in the over all sum there will be units, tens, hundreds and thousands. The required sum .

Answer

Verified

Hint: The question calls for the answer to be without repetition, so we need to solve the sum in that way always reducing the numbers as we go on multiplying. Complete step-by-step answer:
The number of $4-$ digit numbers formed by using 0,2,4,7,8 without repetition ${ }^{5} P_{4}-{ }^{4} P_{3}=120-24=96$
Out of these 96 numbers,
$=545958$ numbers contain 2 in units place.
${ }^{4} P_{3}-{ }^{3} P_{2}$ numbers contain 2 in tens place.
${ }^{4} P_{3}-{ }^{3} P_{2}$ numbers contain 2 in hundreds place.
${ }^{4} P$ numbers contain 2 in units place.
$\therefore$ The values obtained by adding 2 in all numbers.
$$
\left({ }^{4} P_{3}-{ }^{3} P_{2}\right) 2+\left({ }^{4} P_{3}-{ }^{3} P_{2}\right) 20+\left({ }^{4} P_{3}-{ }^{3} P_{2}\right) 200+{ }^{4} P_{2} \times 2000
$$
$={ }^{4} P_{3}(2+20+200+2000)-{ }^{3} P_{2}(2+20+200)$
$=24 \times 2222-6 \times 222$
$=24 \times 2 \times 1111-6 \times 2 \times 111$
Similarly, the value obtained by adding 4 is $24 \times 4 \times 1111-6 \times 4 \times 111$.
The value obtained by adding 7 is $24 \times 7 \times 1111-6 \times 7 \times 111$
The value obtained by adding 8 is $24 \times 8 \times 1111-6 \times 8 \times 111$
Therefore,
The sum of all numbers
$=(24 \times 2 \times 1111-6 \times 2 \times 111)+(24 \times 4 \times 1111-6 \times 4 \times 111)+(24 \times 7 \times 1111-6 \times 7 \times 111)+(24 \times 8 \times 1111-6 \times 8 \times 111)$
$=24 \times 1111 \times(2+4+7+8)-6 \times 111 \times(2+4+7+8)$
$=26664 \times 21-666 \times 21$
$=559944-13986$
$=545958$
Therefore, this is the answer after solving the sum using permutation and combination formulas.

Note: A permutation of an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word "permutation" also refers to the act or process of changing the linear order of an ordered set. Before, solving the sum a student needs to understand the meaning of the word permutation and how to solve them.

2468
2486
2648
2684
2846
2864

8246
8264
8426
8462
8624
+8642

We find that each column has: six 2's, six 4's, six 6's, six 8's.

The total of each column is 6*2 + 6*4 + 6*6 + 6*8 = 120

Hence, the addition has the form:

000120
001200
012000
+120000
------------
133320

Therefore, the sum is: 133,320

Hope I helped : )

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Updated On: 27-06-2022

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