How many ordered triplets (a, b, c), where a, b, and c are positive in

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- The highest number is 8.

Possible triplet (a,b,c) = (8,1,1)
Total arrangements possible \(= \frac{3!}{2!} = 3\)

Case 2- The highest number is 7.

Possible triplet (a,b,c) = (7,2,1)
Total arrangements possible \(= 3! = 6\)

Case 3- The highest number is 6.

Possible triplets (a,b,c) = (6,3,1) and (6,2,2)
Total arrangements possible \(= 3! +\frac{ 3!}{2!} = 9\)

Case 4- The highest number is 5.

Possible triplets (a,b,c) = (5,4,1) and (5,3,2)
Total arrangements possible \(= 3!+3! = 12\)

Case 5- The highest number is 4.

Possible triplets (a,b,c) = (4,4,2) and (4,3,3)
Total arrangements possible \(= \frac{3!}{2!} + \frac{3!}{2!} = 6\)

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

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

A. 28
B. 36
C. 45
D. 54
E. 63

Here 1 ≤ a ≤ 8, 1 ≤ b ≤ 8,1 ≤ c ≤ 8
But 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

Total triplets = \(\frac{8*9}{2} = 36\)

Answer B.
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Thanks to nick1816 and Bunuel , now I can solve these type of problems!
So the restriction is that it is a positive number.

If d=1 and the three positive integers are separated by 2 dashes |
then we have a scenario like this
we have to arrange ddddddd in 3 boxes, boxes are demarcated by 2 dashes
_|_|_

So no. of permutations \(\frac{(7+2)!}{7! 2!} = 36\)

Answer: B

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

If a=1, there are 8 possible triplets: (1,1,8), (1,2,7),...,(1,8,1)
If a=2, there are 7 possible triplets: (2,1,7), (2,2,6),...,(2,7,1)
If a=3, there are 6 possible triplets
...
If a=8, there is only1 possible triplet: (8,1,1)

Thus, there are 8+7+6+...+1=36 ordered triplets that satisfy a+b+c=10

FINAL ANSWER IS (B)

Posted from my mobile device

118=3!/2!=3
127=3!=6
136=6
145=6
226=3
235=6
244=3
334=3

total: 6(4)+3(4)=36

ans (B)

a+b+c=10, below are the values for (a,b,c)

1,1,8
1,2,7
1,3,6
.
.
.
.
1,8,1 = 8pairs


2,1,7
2,2,6
.
.
.
.
.
.
2,7,1= 7pairs

Similarly with
a= 3 we will get 6 pairs
a= 4 we will get 5 pairs
.
.
.
.
a=8 we will get 1 pairs

Total no.of apirs possible is 8+7+6+5+4+3+2+1=36

OA:B

Posted from my mobile device

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

A. 28
B. 36
C. 45
D. 54
E. 63

ordered triplets means (1,8,1) are same as (1,1,8)
a, b, c are natural positive numbers numbers
9c2=36 are the combinations present
Ans B

No. Of positive integral solutions = (n-1) C (r-1)
= 9 C 2 = 9!/7!.2! = 36
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a b and c are positive (atleast 1)

a+b+c = 7

9!/(7!*2!) = 36

B

Bunuel

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