Let $S = {(a, b, c) \in N \times N \times N : a + b + c = 21, a \leq b \leq c}$ and $T = {(a, b, c) \in N \times N \times N : a, b, c are in A. P.}$ , where $N$ is the set of all natural numbers. Then the number of elements in the set $S \cap T$ is

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Solution

The correct option is B $7$
We have $a + b + c = 21 \dots (1)$
Also, $a, b$ and $c$ are in A.P.
Therefore, $b = \frac{a + c}{2} \dots (2)$
From $(1)$ and $(2)$
$b = 7, a + c = 14$
Possible value for $a$ are $1, 2, 3, 4, 5, 6, 7$
Therefore, there are $7$ triplets.

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Similar questions

Q. Let

S = {(a, b, c) \in N \times N \times N : a + b + c = 21, a \leq b \leq c}

and

T = {(a, b, c) \in N \times N \times N : a, b, c a r e i n A P}

, where

N

is the set of all natural numbers. Then, the number of elements in the set

S \cap T

Q. Let

S = {(a, b, c) ϵ N \times N \times N : a + b + c = 21. a \leq b \leq c}

and

T = {a, b, c) ϵ N \times N \times N : a, b, c, a r e i n A . P .}

, where

N

is the set of all natural numbers. Then the number of elements in the set

S \cap T

Q. Consider that

n (S)

represented the number of elements in set S. If

n (A \cup B \cup C) = 40, n (A \cap B^{'} \cap C^{'}) = 5, n (B \cap A^{'} \cap C^{'}) = 10, n (C \cap B^{'} \cap A^{'}) = 6

then number of element which belongs to at least two of the set is

Number of elements in each of the universal set, set A and set B are represented as n(U), n(A), n(B) respectively. The number of elements that are not in set A and not in set B is given by

Number of elements in each of universal set, set A and set B are represented as n(U), n(A), n(B) respectively. The number of elements that are not in set A and not in set B is given by

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Let S={(a,b,c)∈N×N×N:a+b+c=21,a≤b≤c} and T={(a,b,c)∈N×N×N:a,b,c are in A. P.}, where N is the set of all natural numbers. Then the number of elements in the set S∩T is

The correct option is B 7We have a+b+c=21 ⋯(1) Also, a,b and c are in A.P. Therefore, b=a+c2 ⋯(2) From (1) and (2) b=7, a+c=14 Possible value for a are 1,2,3,4,5,6,7 Therefore, there are 7 triplets.

Let $S = {(a, b, c) \in N \times N \times N : a + b + c = 21, a \leq b \leq c}$ and $T = {(a, b, c) \in N \times N \times N : a, b, c are in A. P.}$ , where $N$ is the set of all natural numbers. Then the number of elements in the set $S \cap T$ is

The correct option is B $7$
We have $a + b + c = 21 \dots (1)$
Also, $a, b$ and $c$ are in A.P.
Therefore, $b = \frac{a + c}{2} \dots (2)$
From $(1)$ and $(2)$
$b = 7, a + c = 14$
Possible value for $a$ are $1, 2, 3, 4, 5, 6, 7$
Therefore, there are $7$ triplets.