How many four digit numbers are there such that the digits add up to 10 assuming that all digits are allowed to be zero?

- ParthKohli

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- amistre64

less than 9999 but more than 1000 :)

- amistre64

1117
1171
1711
7111
1126 1162 1261 1621 2611
1216 1612 2161 6121 6211
2116 6112
etc

- ParthKohli

A more mathematical approach, probably?

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## More answers

- mathslover

1000A + 100B + 10C + D = 10
100A + 10B + C + D/10 = 1

- mathslover

if the number is ABCD ^

- amistre64

pfft, that is the most basic mathematical appraoach i know :)

- ParthKohli

I liked what @mathslover said, but that doesn't resolve it. Sorry for taking the medal back. :-(

- mathslover

1000A + 100B + 10C + D = 10 ---- (1)
A + B/10 + C/100 + D/1000 = 1/100 ---- (2)

- ParthKohli

But we must have four equations, no?

- mathslover

No problem :) Parth , thanks though..

- mathslover

I am going to subtract the above 2 eqns

- amistre64

its most likely a counting application with permutations since order matters

- amistre64

how many basic ways are there to define 4 digits that add up to ten, and then how many ways can each group be sorted thru

- ParthKohli

Can we use combinatorics for this?

- mathslover

hmn yes ^^ wait

- amistre64

combinatorics brings to mind the nCr stuff; i think the nPr stuff might be more applicable

- anonymous

sorry to correct what @mathslover is doing is a bit wrong ///////
A+B+C+D=10 is the requirement///////////

- amistre64

1117; how many ways are there to permute this?
1126 how many ways?
1135
1144

- amistre64

we have to account for 0 i spose, which is added computations

- mathslover

OH yes @harsh314 thanks a lot

- amistre64

4+12+12+12
+things with zeros

- amistre64

118: stuff like 1180, 1108, 1018, but not 0118
127
136
145
19
28
37
46
55

- amistre64

so permute the 3 groups and times it by 3

- amistre64

1900, 1090, 1009 looks to be 3 times the 2 groups as well
except for the 55 :) 5500, 5050, 5005 is just 3

- ParthKohli

This is very long. :-|

- ParthKohli

Instead of brute-force, can we try something simpler?

- mathslover

I got it

- amistre64

this isnt brute force anymore, its refining

- mathslover

Lemme check my method again

- mathslover

Forget it . . . I think I must try it again :(

- amistre64

im to focused on my exam in 30 minutes to think thru this clearerer

- mathslover

Parth, do you have answer of it ? ?

- ParthKohli

Nope.

- anonymous

maybe closer the answer

- mathslover

84 numbers ... I think

- ParthKohli

How?

- mathslover

I will put up soln here soon... just thinking of more no.

- mathslover

can the digits repeat Parth?

- ParthKohli

Yes.

- ParthKohli

Well, a number still has four digits if the digits repeat. :-)

- mathslover

I found this on net : http://answers.yahoo.com/question/index?qid=20080330145752AA36Ac2
Now, I am thinking can we use this method to find the answerr?

- ParthKohli

We can think of the ones with zeroes like this:
three digit numbers that add up to 10 * 4!
two digit numbers that add up to 10 * 4!/2
there can't be 3 or 4 zeroes.

- anonymous

5193 is the answer or am i wrong as usual

- ParthKohli

Hey wait, did I mention that the first digit can be zero?

- amistre64

1117 is unique, 4
11xy has 12, and there are 3 ways to chose xy, and no zero groups
4(1)+3(12)
118 is unique, there are 3 ways, and 3 zero groups
1xy has 6 ways, and there are 4 ways to choose xy, and 3 zero groups
3(3)+3(6(4))
55 is unique, theee is ony 1 way, and 3 zero groups
xy 2 ways, there are 4 ways to choose xy, and 3 zero groups
3(1) + 3(2(4))
is there a flaw here?

- mathslover

Oh! Is it allowed to have 0910 as the number ?

- ParthKohli

Yes.

- amistre64

the first digit cant be zero and it still be a 4-digit number tho

- anonymous

no but a number with first digit as 0 does not make sense look at my answer plz i think it is right

- mathslover

^ yeah after that 0910 is not a four digit number , it will be three digit number ..

- amistre64

have fun, i gotta run ;)

- ParthKohli

I know... the question is making such an assumption, so...

- mathslover

@amistre64 ;)

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- mathslover

well there can't be more than 186 numbers.. I think.

- mathslover

^ 196

- mathslover

oh sorry 186

- ParthKohli

Is it 186?

- anonymous

more than 577

- mathslover

yes it is 186 @ParthKohli

- ParthKohli

The answer is 186?

- anonymous

(0..9).each do |a|
(0..9).each do |b|
(0..9).each do |c|
(0..9).each do |d|
count = count + 1 if (a + b + c + d) == 10
end
end
end
end
puts count
gave me 282 as answer, I don't know a mathematical approach though :)

- mathslover

yes @ParthKohli

- ParthKohli

186 when 0 is allowed or not?

- mathslover

allowed

- ParthKohli

How? :S

- ParthKohli

282 and 186 are both not right :-|

- anonymous

i think i got the answer.............

- anonymous

282 was with leading zeros, without it would be 219

- anonymous

408 without the zeros

- anonymous

468 with zeroes

- anonymous

plz do reply if wrong.......

- anonymous

@ParthKohli

- ParthKohli

I'd check

- ParthKohli

Done! 13 C 3 = 286

- anonymous

yes i got this too a silly error how did you do it............

- anonymous

my method is tooo long.......

- anonymous

@ParthKohli

- ParthKohli

... 286 is the answer

- anonymous

i too got the same but by long method how did you get it .........

- anonymous

I still find it odd that my bruteforce method which calculates the sum of the digits from 0 upto 9999 is off by 4

- ParthKohli

We wanted the number of non negative numbers \(a_1,a_2,a_3,a_4\) such that \(a_1 + a_2 + a_3 + a_4 = 10\). A formula says that it would be \(\binom{10 + 4 - 1}{4 - 1}\)

- anonymous

oh so no logic behind it as such..............

- anonymous

But bruteforcing every digit from 0 to 9999 gives 282 though, shouldn't they be the same?

- ParthKohli

I don't know. :-|

- anonymous

Weird :l

- anonymous

my method was long and a bit calculation mistake..............
case1:::'1" as the first digit ,,,,,, the following subcases arise
(a) .....with 2 zeroes ........9 has to be the sum of the remaining three numbers so total possible combinations are 3.
(b) .......with 1 zero we have 810,720,630,540......all these have 6 combinations each so total 24
(c) .......with no zeroes.....711(3combinations),621(6),531(6),522(3),432(6),441(3),333(1)
so total of 28 combinations
hence for the 1st case we have 45 combinations similarly we can have others tooo..........and then add all........we get 286

- anonymous

but it would be great if any einstein of the present century could explain how @ParthKohli 's method came into being.............

- anonymous

@satellite73

- amistre64

works in google chrome :)

0000:

1000:

0000:

1000:

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- amistre64

i should prolly include an output such that we can verify its a+b+c+d parts to make sure its operating as expected

- anonymous

I'm working on one :)

- amistre64

gotta weed out a bug :)
0,0,0,10
0,0,1,9
0,0,2,8
0,0,3,7
0,0,4,6
0,0,5,5
0,0,6,4
0,0,7,3
0,0,8,2
0,0,9,1
0,0,10,0
0,1,0,9
0,1,1,8

- amistre64

if abcd = 10 then reset to 0, not if greater than 10 .... lol

- amistre64

im missing 9001, 9010, 9100 from the end of my 0xxx for some reason
279+3 = 282
otherwise its 219 if we use actual 4-digit numbers :)

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- anonymous

282 is correct :)

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- amistre64

yay!! :)

- amistre64

... brute math, the easy way :)

- anonymous

Computers made these problem a lot easier :)

- anonymous

Unfortunately Parth has 286 as answer though :(

- amistre64

parth is counting somethings more than once then.
i adjusted my loops to 0 to 10000 and it caught the last 3 ;)

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