Question

how many triples of natural numbers have sum of 11

Answer 1

is it 13!/(11!*2!) -12!*2!/(11!*2!)= 78-12=66

Answer 2

66

Answer 3

is it 13!/(11!*2!) -12!*2!/(11!*2!)= 78-12=66

Answer 4

because the numbers must be natural numbers, so the sum of triples are :
(11-1)C(3-1) = 10C2 = 45

Answer 5

u can re-write in (11+3-2)C(3-1) =66

Answer 6

i have tried by manual (exploration method) too, it can be 45, @matricked
if im not mistake

Answer 7

@RadEn
can u explain...

Answer 8

i want to answering by exploration
let the numbers are a,b, and c (with a,b, and c are natural numbers)
so, we get an equation a+b+c=11

* case I, if a=1 then b+c=10 or c=10-b
to get c is natural number, so the value of b must be = 1,2,3,..,9
therefore there are 9 order pairs (a,b,c)

* case II, if a=2 then b+c=9 or c=9-b
to get c is natural number, so the value of b must be = 1,2,3,..,8
therefore there are 8 order pairs (a,b,c)
....
so on, until :
* case 9, if a=9 then b+c=2
just 1 option only, it satisfied for b=c=1
therefore there is 1 order pair (a,b,c)

so, the total order pairs (a,b,c) = 9+8+7+6+...+2+1 = 45

Answer 9

am i right, @matricked ?

Answer 10

@RadEn
yup dear you are absolutely right..
i have aslo verified the same by writing a simple program and found out to be 45

i have also include the ways in which a=0 (not a natural number ) and c=0 (not a natural number and hence it will be 66 -(10+10+1)=45 ways
ur explaination is also very clear
thank you very much for correcting me
@choze
my ans isn't needs correction
so follwo RadEn's method...