Latest Elitmus Aptitude Question SOLUTION: If a+b+c=21, what is the total number of non-negative integral solutions?

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(#M40155284) ELITMUS Numerical Ability question PERMUTATIONS & COMBINATIONS Keep an EYE

If a+b+c=21, what is the total number of non-negative integral solutions?

Asked In Elitmus Bhavya Dhawal (8 years ago)

Solved Amit Sankhyan Read Solution (10)

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Read Solution (10) :

Amit Sankhyan  (8 years ago) Best Solution

Easy Way. Think of it is a problem of diving 21 items to three people A, B & C which may or may not receive an item.

So simply use Combination here as (n-r+1) C (r-1) i.e.

(21+3-1) C (3-1) => 23 C 2 == > 253 solutions.

Is this solution Helpfull? Yes (27) | No (6)  

Rounak Shah  (8 years ago)

divide 21 items to 3 people A, B & C which may or may not receive an item.
So simply use Combination here as (n+r-1) C (r-1) i.e.

(21+3-1) C (3-1) => 23 C 2 == > 253 solutions.

Is this solution Helpfull? Yes (9) | No (1)  

rahul  (8 years ago)

see this link, u get a clear idea abt it!!!
http://www.careerbless.com/aptitude/qa/permutations_combinations_imp8.php

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Bhavya Dhawal  (8 years ago)

253 is the right ans wer but I didnt understand the solution.

Is this solution Helpfull? Yes (4) | No   

Rupa Kumari  (8 years ago)

Ans= 28

fix value of a from 3 to 9 such as:
if a=3; then b+c=18; =) b=c=9 i.e. 1 way
if a=4; then b+c=17; =) b=8/9; c=8/9 i.e. 2 ways

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if a=9; then b+c=12;

you will get total of 28 non-negative integral solutions

Is this solution Helpfull? Yes (1) | No (4)  

Lavanya  (8 years ago)

22C2
=>231

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siddharthjain111@gmail.com  (8 years ago)

We can also solve this through partition method (this is method is used for finding the no. of solutions or distribution of identical things :
a + b + c = 21

the question said that we have separate this value among 3 (a, b, c)

then,

a + b + c =21!+2p! : p=partition ( we take 2p because we have to separate 21 in three(a , b , c)).

a + b + c = 23! / 21! * 2! = 253 answer.

(It is much simpler rather to remembering the formulae).

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Rahul  (8 years ago)

Since, a+b+c=21
Now for integral solution a>=1;b>=1;c>=1
therefore, a=1+i1;b=1+i2;c=1+i3;
again, i1+i2+i3=21-3=19=n;r=3
using, (n+r-1)C(r-1)=253

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AZEEM KHAN  (8 years ago)

22+21+20+........+2+1 = 253

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Sanjay Singh  (8 years ago)

(n+r-1)C(r-1). Here, n=21 and r=3.

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