Question: How many ordered triplets (a, b, c), where a, b, and c are positive integers, are there such a + b + c = 10 ?

1. 28
2. 36
3. 45
4. 54
5. 63

Correct Answer: B
Solution and Explanation:
Approach Solution 1:

The problem statement suggests that:

Given:

• a + b + c = 10

Find out:

• The number of ordered triplets (a, b, c), where a, b, and c are positive integers.

a + b + c = 10

\(\frac{a + b + c}{3} = 3.33\)

The highest among them can take 5 values(4, 5, 6, 7 and 8)

Case 1: When the highest number is 8, then possible triplet (a,b,c) = (8,1,1)
Total arrangements possible = 3!/2! = 3

Case 2: When the highest number is 7, then possible triplet (a,b,c) = (7,2,1)
Total arrangements possible = 3! = 6

Case 3: When the highest number is 6, then possible triplets (a,b,c) = (6,3,1) and (6,2,2)
Total arrangements possible = 3! + 3!/2! = 9

Case 4: When the highest number is 5, then possible triplets (a,b,c) = (5,4,1) and (5,3,2)
Total arrangements possible = 3! + 3! = 12

Case 5: When the highest number is 4, then possible triplets (a,b,c) = (4,4,2) and (4,3,3)
Total arrangements possible = 3!/2! + 3!/2! = 6

Number of ordered triplets (a, b, c), where a, b, and c are positive integers, are there such “a + b + c = 10” = 3 + 6 + 9 + 12 + 6 = 36

Approach Solution 2:

The problem statement states that:

Given:

• a + b + c = 10

Find out:

• The number of ordered triplets (a, b, c), where a, b, and c are positive integers.

Since a + b + c = 10, then we can say,
Here, 1 ≤ a ≤ 8, 1 ≤ b ≤ 8, 1 ≤ c ≤ 8
However, when
a = 1; 1 ≤ b ≤ 8, 1 ≤ c ≤ 8 --- 8
a = 2; 1 ≤ b ≤ 7, 1 ≤ c ≤ 7 --- 7
a = 3; 1 ≤ b ≤ 6, 1 ≤ c ≤ 6 --- 6
a = 4; 1 ≤ b ≤ 5, 1 ≤ c ≤ 5 --- 5
a = 5; 1 ≤ b ≤ 4, 1 ≤ c ≤ 4 --- 4
a = 6; 1 ≤ b ≤ 3, 1 ≤ c ≤ 3 --- 3
a = 7; 1 ≤ b ≤ 2, 1 ≤ c ≤ 2 --- 2
a = 8; b = 1, c = 1 --- 1

Therefore, total triplets = \(\frac{8∗9}{2} = 36.\)

Approach Solution 3:

The problem statement states that:

Given:

• (a, b, c) are positive integers where a+b+c=10

Find out:

• The number of ordered triplets (a, b, c).

Since a+b+c=10 and a, b, c are positive integers, we can say:

No. of positive integral solutions = (n-1) C (r-1) = 9 C 2 = 9! /7!.2! = 36

“How many ordered triplets (a, b, c), where a, b, and c are positive integers”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. The candidates can enhance their skills by practising more GMAT Problem Solving questions. The GMAT Quant practice papers help the candidates to become familiar with varieties of questions that will further improve their mathematical understanding.

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