M12-12

Question

If  x ,  y , and  z  are positive integers, how many different ordered triplets  (x, y, z)  are possible such that  x + y + z = 5 ?

A. 3
B. 4
C. 5
D. 6
E. 8

Bunuel · Answer

Official Solution:

If  x ,  y , and  z  are positive integers, how many different ordered triplets  (x, y, z)  are possible such that  x + y + z = 5 ?

A. 3
B. 4
C. 5
D. 6
E. 8

There are two ways to break the number 5 into the sum of three  positive integers :  2+2+1  and  3+1+1 . Since we are interested in  ordered  triplets, each combination can be represented by three distinct triplets:  (2, 2, 1) ,  (2, 1, 2) ,  (1, 2, 2)  and  (3, 1, 1) ,  (1, 3, 1) ,  (1, 1, 3) , respectively.

Answer: D

abhilash53 · Answer

Can this problem be solved using the combinations formula?

aparajita123 · Answer

well try this

to get 5 we have two combination of 3 integers

1- 1,2,2
2- 1,1,3

we can arrange option 1 - 1,2,2 in 3!/2! ways = 3
same way for option 2 - 1,1,3 in 3!/2! ways = 3

3+3 = 6

( 3! => because the set consists of 3 numbers ,,, 2! because two of the 3 numbers are identical)

lillianmbula · Answer

Bunuel how did you come to that conclusion? I thought the x and y and z have to be different numbers?

Bunuel · Answer

Unless it is explicitly stated otherwise, different variables CAN represent the same number.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Shaurs · Answer

I don’t quite agree with the solution. why is ( 0, 2 , 3 ) and ( 0, 1 , 4 ) not considered ? If they are not ordered triplets then why ( 1, 1, 3 ) is considered as a ordered triplet

Bunuel · Answer

Check the highlighted part in the question:

If  x ,  y , and  z  are   positive   integers, how many different ordered triplets  (x, y, z)  are possible such that  x + y + z = 5 ?

0 is not a positive number.

Sushi_545 · Answer

Hi  Bunuel ,

here no is 5 if no is large its difficult to find out pair can we use this method and concept?:-

If the number of non negative integer solutions for the equation x1+x2 + ..+ xn = nx1+x2 + ..+ xn = n, then the number of ways the distribution can be done is:

(i) n+r−1Cr−1. In this case, value of any variable can be zero.

(ii) n−1Cr−1. In this case, minimum value for any variable is 1.

here applying second case since min value should be one we have n=5, r=3

4C2=6

M12-12

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