Latest TCS Aptitude Question SOLUTION: a,b,c,d are positive integers.find the no of possible solutions to the equation a+b+c+d=10

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a,b,c,d are positive integers.find the no of possible solutions to the equation a+b+c+d=10

Asked In TCS karthik (10 years ago)

Unsolved Read Solution (18)

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Read Solution (18) :

Tanu  (10 years ago)

Its the direct application of Stars & bars Rule.
Now putting a=a-1,b=b-1,c=c-1,d=d-1
The above equation reduce to a+b+c+d=10-4=6
a+b+c+d=6
Now there are 6 stars and 4-1(a,b,c,d) bars
So the combination
n+k-1 C k-1 = 6+(4-1) C 4-1
= 9 C 3
= 84

Is this solution Helpfull? Yes (42) | No (6)  

Rajarshi Bhattacharyya  (10 years ago)

13C3 Formula : (n+r-1)C(r-1)

Is this solution Helpfull? Yes (13) | No (5)  

NARAYANA SWAMY DAMMALA  (10 years ago)

IF given integers are whole numbers ans is 13c3
if natural numbers ans is 9c3

Is this solution Helpfull? Yes (13) | No (1)  

satheesh  (10 years ago)

using following formula "(n+r-1)c(r-1)" to find out the result
here r=4(a,b,c,d) and n=10......so
subsuit (10+4-1)c(4-1)
13c3=286
so answer is 286

Is this solution Helpfull? Yes (7) | No (6)  

shubh mehta  (10 years ago)

7,1,1,1 = 4 cases
6,2,1,1 = 12 cases
5,3,1,1 =12 cases
5,2,2,1 = 12 cases
4,3,2,1 = 24 cases
4,4,1,1 = 6 cases
4,2,2,2= 4 cases
74 ans

Is this solution Helpfull? Yes (6) | No (9)  

Bholanath   (10 years ago)

9C3.......

Is this solution Helpfull? Yes (5) | No (2)  

nitin garg  (10 years ago)

9c3
a,b,c,d are 4(a) terms...
n we have to make der some 10(n)...
so
(n-1)c(a-1) will give you answer

Is this solution Helpfull? Yes (2) | No (2)  

Neelam Khilwani  (10 years ago)

a+b+c+d=10 where a,b,c,d can even take whole no values.so to ease at taking natural no values we do:(a+1)+(b+1)+(c+1)+(d+1)=10+4
W+X+Y+Z=14(is the same as previous,but here X,Y,Z and W take only natural no solutions)
let us now take 14 as 14 empty seats and since the values of X,W,Y,z changes every instant,we can take operator(+) as the no of persons ,but since the operators to be put in between operands,we get 13 vacant places and these 3 are put in 13c3 ways in them
eg: _ _ _ _ _ _ _ _ _ _ _ _ _ (spaces) + + + (operators) can be put in between in 13c3

Is this solution Helpfull? Yes (1) | No (3)  

pre  (10 years ago)

you can easily solve this question->
since it is given that a, b,c,d are positive integers so we apply the formula n-1Cr-1 instead of n+r-1Cr-1 bcoz this formula is for non negative integers (including 0) hence answer is 9C3

Is this solution Helpfull? Yes (1) | No   

mukesh  (10 years ago)

no sol
bcoz there r 4 variabls...?

Is this solution Helpfull? Yes (1) | No (1)  

moumita ghose  (10 years ago)

10+4-1C(4-1)=7C3=35

Is this solution Helpfull? Yes | No (6)  

Saptarsi Basu  (10 years ago)

If (a+b)=x ; (c+d)=y ; then the above equation can be represented as
x+y=10 -----> This gives us Infinite solutions.
Now If we consider a,b,c,d and equation a+b+c+d=10
by considering the possible cases, this gives us 74 Positive solutions and
78 Non-Negative Solutions.

Is this solution Helpfull? Yes | No (3)  

nisha sengar  (10 years ago)

sateesh pls explain whyu hv used that formula

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shivam khanduja  (10 years ago)

24 possible solutions.
because 1+2+3+4=10 is the only possible situation as a,b,c,d are different positive integers.
1,2,3,4 can be arranged in 4! ways.hence 24 solutions.

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Anandakumar k  (10 years ago)

plz...friends explain again....

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Sonali Agarwal  (10 years ago)

24 solutions are there as
1+2+3+4,1+3+2+4,1+4+2+3,1+3+4+2,1+2+4+3, and 1+4+3+2
for 1 integer we have 6 solutions so for 4 integers it is 6*4 i.e, 24 solutions.

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vignesh  (10 years ago)

formula
"(n+r-1)c(r-1)"
to find result
here r=4(a,b,c,d) and n=10......so
substitute (10+4-1)c(4-1)
13c3=286

286 is the answer

Is this solution Helpfull? Yes | No (1)  

Venky Cherukuri  (9 years ago)

1+2+3+4 = 10
So, total 4! possible no.of solutions available.

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