Question

How many 3-digit numbers are there where the sum of the digits equals 11?

Answer 1

Any ideas? Can you come up with maybe a specific answer that works? This seems to be closely related to how many ways can you arrange 3 numbers.

Answer 2

I wrote out all the combinations (1+1+9, 1+2+8, etc.) which therefore gives 9 possibilites, and multiplied it by 6, because you can invert a 3 digit number 6 ways.

Answer 3

are you familiar with stars and bars ?

Answer 4

no @ganeshie8

Answer 5

stars and bars directly gives you the answer, but lets work it by counting :

say the 3 digit number is \(abc\), such that :
\[a+b+c = 11 \\ 0\lt a \le 9\\0\le b\le 9\\ 0\le c \le 9\]

Answer 6

Ok.

Answer 7

Honestly I don't know what that means at all, sorry

Answer 8

@ganeshie8

Answer 9

lets say a = 1, then we get :
\[\large b+c = 10\]

can you guess how many ways you can cook up b and c ?

Answer 10

isn't it 5?

Answer 11

whats the minimum possible value of \(b \) ?

Answer 12

1

Answer 13

for sure, \(b\) cannot be 0 (why ?)
\(b\) has to range from 1 to 9
and for each of this value of b, c us uniquely determined. so total number of possible ways for b+c=10 is 9

Answer 14

oh yeah because you can swap them.

Answer 15

Exactly ! corrected some typoes :

for sure, \(b\) cannot be \(0\) (why ?)
\(b\) has to range from \(1\) to \(9\)
and for each of this value of \(b\), the value of \(c\) can be uniquely determined.
so total number of possible ways for \(b+c=10\) is \(9\)

Answer 16

yep.

Answer 17

a=1 : you get 9 possible numbers
can you work a=2 ?

Answer 18

8

Answer 19

hmm lets see...

when a = 2 :
\[\large b+c = 9\]

whats the minimum value of \(b\) ?
whats the maximum value of \(b\) ?

Answer 20

maximum 8 min 1

Answer 21

think again
why can't the value of \(b\) be 0 or 9 ?

Answer 22

oh yeah!

Answer 23

a=2, b+c=9
gives you 10 numbers

Answer 24

we can sit and count all remaining cases (a=3,4,5,6,7,8,9)
or look for a pattern...

Answer 25

lets work a=3 case, then we will see a pattern may be.. .

Answer 26

ok, there are 7 possibilites.

Answer 27

actually aren't there nine? because 0 and 8 can be used.

Answer 28

exactly ! looks we have got a pattern :3

Answer 29

ok, so for every possibility of a there are nine possibilities for b and c?

Answer 30

a=1 : b+c=10 ( 9 possible numbers)
a=2 : b+c= `9` ( `10` possible numbers)
a=3 : b+c= `8` ( `9` possible numbers)
a=4 : b+c= `7` ( `8` possible numbers)
....

Answer 31

^^

Answer 32

oh ok.

Answer 33

for a=8, b+c = `3`, you will get `4` possible numbers

Answer 34

In general :
the equation \(b+c = n\) has \(n+1\) possible non negative solutions

Answer 35

yes, except for the first one.

Answer 36

yes :) thats because we cannot allow "10" for b or c
so whats the final answer ?

Answer 37

61!

Answer 38

So how is this related to the equation that you were talking about before?

Answer 39

thats just a lazy way of expressing the sum :
9 + 10 + 9 + 8 + ... + 3

right ?

Answer 40

Yes.

Answer 41

So how do u figure it though?

Answer 42

above sum be represented as below :

\[\large \sum \limits_{a=1}^9 ((11-a) +1) - 2\]

Answer 43

How would I get that in the first place?

Answer 44

that comes from solving a simpler equation : \(\large x+y = n\)

Answer 45

how many solutions exist in non negative integers ?

Answer 46

the answer is \(\large n+1\), right ?

Answer 47

yes, but we didn't know that before we figured out the answer.

Answer 48

thats right ! based on that can u cookup a formula for number of solutions of \(\large x+y \le n\) ?

Answer 49

above formula is almost same as the question you have originally posted (a+b+c=11)

Answer 50

i don't understand what the variables represent

Answer 51

what variables ?

Answer 52

x, y and n

Answer 53

okay, let me give a specific value for \(n\) and rephrase the problem :

How many "non negative integer solutions" are there for the equation \(\large x+y \le 10\) ?

Answer 54

ok.

Answer 55

I'll give you the answer, try the proof in ur free time :)

Answer : \(\large \sum \limits_{k=0}^{10} (k+1
)\)

Answer 56

what does the e mean

Answer 57

what e ?

Answer 58

the equation sign

Answer 59

\[\large \sum \limits_{k=0}^4 k \]

is just a lazy way of writing :

\[\large 0 + 1+2+3+4\]

Answer 60

this site explains that SUM notation better :
http://www.mathsisfun.com/algebra/sigma-notation.html

Answer 61

what is he k on the right

Answer 62

oh ok i understand now.

Answer 63

so how in the first place did u formulate the equation

Answer 64

good :)

Answer 65

Spoiler : proof @
http://openstudy.com/users/ganeshie8#/updates/53cbc2dee4b0bff525b1e8f1

Answer 66

isn't that just your updates page?

Answer 67

so i don't understand how you made the equation @ganeshie8

Answer 68

scroll down a bit to see the actual proof

Answer 69

ok thank you very much @ganeshie8

Answer 70

np :) let me know if something is not clear in that page