Find the sum of all 4-digit numbers formed by taking all the digits 2, 4, 5, and 6.

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.

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Manager

Joined: 09 May 2009

Posts: 124

Find the sum of all the four digit numbers which are formed [#permalink]

  Updated on: 06 Nov 2012, 02:22

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Question Stats:

71% (01:38) correct 29% (01:51) wrong based on 544 sessions

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Find the sum of all the four digit numbers which are formed by digits 1, 2, 5, 6

A. 933510
B. 93324
C. 65120
D. 8400

Originally posted by xcusemeplz2009 on 23 Dec 2009, 05:38.
Last edited by Bunuel on 06 Nov 2012, 02:22, edited 1 time in total.

Renamed the topic and edited the question.

Math Expert

Joined: 02 Sep 2009

Posts: 87599

Re: permutation [#permalink]

  23 Dec 2009, 06:16

xcusemeplz2009 wrote:

find the sum of all the four digit numbers which are formed by digits 1,2,5,6

a)933510
b)93324
c)65120
d)8400

The answer choices make the solution easy:

There are 4!=24 four digit numbers which are formed by digits 1, 2, 5, 6.

Obviously 24/4=6 numbers will end with 1; 6 numbers will end with 2, 6 numbers with 5 and 6 numbers with 6.

6*1+6*2+6*5+6*6=6*14=84, which means that the sum of all these 24 numbers must end by 4, only answer choice with 4 at the end is B.

Answer: B.

But if we were not given such an easy answer choices, the solution would be:

We have 24 numbers of the of the form: 1000a+100b+10c+d, where a, b, c and d can take any value from the set {1, 2, 5, 6} and there will be 6 numbers with same digit (a, b, c, d) at the thousands, hundreds, tens and units digits.

(1000*6a+1000*6b+1000*6c+1000*6d)+(100*6a+100*6b+100*6c+100*6d)+(10*6a+10*6b+10*6c+10*6d)+(6a+6b+6c+6d)=(6a+6b+6c+6d)(1111)=
=6(a+b+c+d)(1111)=6*14*1111=93324

Generally the sum of all the numbers which can be formed by using the n distinct digits, is given by the formula:

(n-1)!*(sum of the digits)*(111…..n times)
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Re: Find the sum of all the four digit numbers which are formed [#permalink]

  03 Oct 2017, 22:08

Doesn't 'sum of all' suggest that repetition is allowed ?

VP

Joined: 06 Sep 2013

Posts: 1379

Concentration: Finance

Re: Find the sum of all the four digit numbers which are formed [#permalink]

  30 Dec 2013, 07:27

xcusemeplz2009 wrote:

Find the sum of all the four digit numbers which are formed by digits 1, 2, 5, 6

A. 933510
B. 93324
C. 65120
D. 8400

3!*1111*(14) will have units digit 4

B is the answer

Hope it helps
Cheers!
J

Intern

Joined: 20 Sep 2015

Posts: 14

Location: India

Re: Find the sum of all the four digit numbers which are formed [#permalink]

  11 Oct 2017, 18:45

zvazviri wrote:

Doesn't 'sum of all' suggest that repetition is allowed ?

I have the same query. Experts, please clarify.

Math Expert

Joined: 02 Sep 2009

Posts: 87599

Re: Find the sum of all the four digit numbers which are formed [#permalink]

  11 Oct 2017, 20:00

gumnamibaba wrote:

zvazviri wrote:

Doesn't 'sum of all' suggest that repetition is allowed ?

I have the same query. Experts, please clarify.

No it does not. Ideally it should have been clarified more precisely but it's assumed that we are adding the numbers which could be constructed by re-arranging the numbers 1, 2, 5, and 6.
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Re: Find the sum of all the four digit numbers which are formed [#permalink]

  11 Oct 2017, 22:23

Bunuel wrote:

gumnamibaba wrote:

zvazviri wrote:

Doesn't 'sum of all' suggest that repetition is allowed ?

I have the same query. Experts, please clarify.

No it does not. Ideally it should have been clarified more precisely but it's assumed that we are adding the numbers which could be constructed by re-arranging the numbers 1, 2, 5, and 6.

Given this ambiguity, it's safe to say this is not an OG problem, and I would not see problems worded similarly on the test?

Math Expert

Joined: 02 Sep 2009

Posts: 87599

Re: Find the sum of all the four digit numbers which are formed [#permalink]

  11 Oct 2017, 22:26

zvazviri wrote:

Given this ambiguity, it's safe to say this is not an OG problem, and I would not see problems worded similarly on the test?

Yes, this is not a proper GMAT problem not only because of the wording. Notice that it has 4 options not 5.
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Find the sum of all the four digit numbers which are formed [#permalink]

  24 Nov 2018, 03:37

Could also be solved with estimation:

\(4*1000 + 4*2000 + 4*5000 + 6*6000=68000\) Just looking at the answer choices option B is the only possible solution.
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Re: Find the sum of all the four digit numbers which are formed [#permalink]

  24 Nov 2018, 09:56

xcusemeplz2009 wrote:

Find the sum of all the four digit numbers which are formed by digits 1, 2, 5, 6

A. 933510
B. 93324
C. 65120
D. 8400

1256+6521=7777
7777/2=3888.5
3888.5*4!=93324
B

Director

Joined: 08 Aug 2017

Posts: 728

Re: Find the sum of all the four digit numbers which are formed [#permalink]

  19 Sep 2019, 08:32

I have made similar mistakes in other problems.
Until repetition is not specified in problem, it is safe to proceed considering non-repetition of digits/letters.
And this rule is valid throughout the GMAT.

Bunuel wrote:

gumnamibaba wrote:

zvazviri wrote:

Doesn't 'sum of all' suggest that repetition is allowed ?

I have the same query. Experts, please clarify.

No it does not. Ideally it should have been clarified more precisely but it's assumed that we are adding the numbers which could be constructed by re-arranging the numbers 1, 2, 5, and 6.

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Location: Malaysia

Concentration: Strategy, General Management

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Find the sum of all the four digit numbers which are formed [#permalink]

  28 Jan 2020, 22:37

Hi Bunuel , thanks for the good explanation as always. Regarding the formula below, can you confirm my understanding that it is not applicable if there is a "0" in one of the digits provided?

Based on a similar question from the link below, the calculation for this question can also be done as:
Total possible 4-digit numbers= 4*3*2*1= 24

Sum of digits at unit place= 24/4*(1+2+5+6)=84
Sum of digits at tens place= 24/4*(1+2+5+6)=84
Sum of digits at hundreds place= 24/4*(1+2+5+6)=84
Sum of digits at thousands place= 24/4*(1+2+5+6)=84

the sum of all possible 4 digit numbers that can be formed using all the digits of the number 1256
= 84* (1000+100+10+1)=93,324

https://gmatclub.com/forum/find-the-sum ... s#p2448163

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Posts: 6883

Location: Canada

Re: Find the sum of all the four digit numbers which are formed [#permalink]

  04 Sep 2022, 12:09

xcusemeplz2009 wrote:

Find the sum of all the four digit numbers which are formed by digits 1, 2, 5, 6

A. 933510
B. 93324
C. 65120
D. 8400

There are 4 ways to select the thousands digit
There are 4 ways to select the hundreds digit
There are 4 ways to select the tens digit
There are 4 ways to select the ones digit
So, the total number of 4-digit numbers possible = (4)(4)(4)(4) = 256

Now let's focus on the thousands digits of our 256 numbers.
64 (aka 1/4) of the 256 numbers will have thousands digit 1. 64 x 1000 = 64,000
64 (aka 1/4) of the 256 numbers will have thousands digit 2. 64 x 2000 = 128,000
64 (aka 1/4) of the 256 numbers will have thousands digit 5. 64 x 5000 = 320,000
64 (aka 1/4) of the 256 numbers will have thousands digit 6. 64 x 6000 = 384,000
64,000 + 128,000 + 320,000 + 384,000 = 896,000

So, if we ignore the hundreds, tens and ones digits, the sum of our 256 numbers is already 384,000, which means the TOTAL sum must be greater than 384,000

Answer: A
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Re: Find the sum of all the four digit numbers which are formed [#permalink]

04 Sep 2022, 12:09

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