edited by
3,333 views

3 Answers

Best answer
10 votes
10 votes
We have 17 and we want to distribute it to x, y and z such that each gets at least one. So, initially itself we take 3 and give 1 each to x, y, and z. So, now we have 14 left and have to divide this into 3. Let the 14 be as follows and we use | for the division.

X X X X X X X X X X X X X X

We need 2 |  for dividing this into 3 parts. and with 2 |, our problem reduces to finding all the permutations of 14 + 2 items, where 14 and 2 of them are identical. So, our answer will be

$16!/14!2! = 15 * 8 = 120$
edited by
6 votes
6 votes
Consider that x=1, then it'll have 15 different solutions in positive integers. (1,1,15), (1,2,14) ........(1,15,1).

Similarly consider x=2, then it'll have 14 different solutions in positive integers.

 

Therefore, total number of different solutions in positive integers = 15 + 14 +  ....... +1

                                                            = 120 solutions

Answer would be (A).
edited by
2 votes
2 votes

Based on standard formula:

This problem is based on standard results: 

No. of non-negative solution for equation x1+x2+x3+-----+xn = r   ,where x1,x2...xn>=0    is   n+r-1Cr.

Given eqn:  x1 + x2 + x3 = 17, here x1,x2,x3>=1. ----------eq(A)

Converting this eqn in standard form as   (x1-1) + (x2-1) + (x3-1) = 14  .

Let a = x1-1; b=x2-1; c= x3-1; hence a,b,c >=0.So eqn reduces to   a+b+c = 14 where a,b,c>=0 --------eq(B)

Hence no. of positive integral sol for eq(A) is equivalent to no.of non-negative solutions for eq(B).

Hence reqd  no. of solutions = n+r-1C = 3+14-1C14 = 16C14 16C= 120  Ans.

P.S: This problem is subset of more generic problem a1x1+a2x2+a3x3+----------------anxn  <= r where ki <= ai <= ki  

Related questions

1 votes
1 votes
2 answers
1
1 votes
1 votes
3 answers
3
0 votes
0 votes
3 answers
4
snaily16 asked Jan 22, 2019
1,373 views
The number of ways 5 letter be put in 3 letter boxes A,B,C. If letter box A must contain at least 2 letters.