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Count A.P. Triplets with a+b+c=21 and a≤b≤c | JEE

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Question
Let S={(a,b,c)∈N×N×N:a+b+c=21, a≤b≤c}S = \{(a, b, c) \in N \times N \times N : a + b + c = 21,\ a \le b \le c\} and T={(a,b,c)∈N×N×N:a,b,c are in A.P.}T = \{(a, b, c) \in N \times N \times N : a, b, c \text{ are in A.P.}\}, where NN is the set of all natural numbers. What is the number of elements in S∩TS \cap T?
A66
B77correct
C1313
D1414
Solution
Step 1: a,b,ca, b, c in A.P. ⇒2b=a+c\Rightarrow 2b = a + c. With a+b+c=21a + b + c = 21,
3b=21⇒b=7,a+c=14.3b = 21 \Rightarrow b = 7, \qquad a + c = 14.
 Step 2: a≤b≤c⇒a≤7≤ca \le b \le c \Rightarrow a \le 7 \le c, with a∈Na \in N and c=14−ac = 14 - a:
a=1,2,3,4,5,6,7⇒c=13,12,11,10,9,8,7.a = 1, 2, 3, 4, 5, 6, 7 \Rightarrow c = 13, 12, 11, 10, 9, 8, 7.
 ∴S∩T\therefore S \cap T has 77 elements. Correct answer: (2)
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