JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 18)
Let $$x$$ and $$y$$ be distinct integers where $$1 \le x \le 25$$ and $$1 \le y \le 25$$. Then, the number of ways of choosing $$x$$ and $$y$$, such that $$x+y$$ is divisible by 5, is ____________.
Answer
120
Explanation
Let $x+y=5 \lambda$
Possible cases are
$\begin{array}{llc}x & y & \text { Number of ways } \\ 5 \lambda(5,10,15,20,25) & 5 \lambda(5,10,15,20,25) & 20 \\ 5 \lambda+1(1,6,11,16,21) & 5 \lambda+4(4,9,14,19,24) & 25 \\ 5 \lambda+2(2,7,12,17,22) & 5 \lambda+3(3,8,13,18,23) & 25 \\ 5 \lambda+3(3,8,13,18,23) & 5 \lambda+2(2,7,12,17,22) & 25 \\ 5 \lambda+4(4,9,14,19,24) & 5 \lambda+1(1,6,11,16,21) & 25\end{array}$
Total number of ways $=20+25+25+25+25=120$
Note : In first case total number of ways = 20 as in the question given that the chosen $$x$$ and $$y$$ are distinct integers each time. So when you choose x as 5 then you can't choose y as 5. Possible values of y are (10, 15, 20, 25). So, here four possible pairs of (x, y) possible { (5, 10), (5, 15), (5, 20), (5, 25)}.
Similarly, four possible pairs of (x, y) possible each time when x = 10, 15, 20 and 25.
$$ \therefore $$ Total number of ways in the first case = 5 $$ \times $$ 4 = 20.
Possible cases are
$\begin{array}{llc}x & y & \text { Number of ways } \\ 5 \lambda(5,10,15,20,25) & 5 \lambda(5,10,15,20,25) & 20 \\ 5 \lambda+1(1,6,11,16,21) & 5 \lambda+4(4,9,14,19,24) & 25 \\ 5 \lambda+2(2,7,12,17,22) & 5 \lambda+3(3,8,13,18,23) & 25 \\ 5 \lambda+3(3,8,13,18,23) & 5 \lambda+2(2,7,12,17,22) & 25 \\ 5 \lambda+4(4,9,14,19,24) & 5 \lambda+1(1,6,11,16,21) & 25\end{array}$
Total number of ways $=20+25+25+25+25=120$
Note : In first case total number of ways = 20 as in the question given that the chosen $$x$$ and $$y$$ are distinct integers each time. So when you choose x as 5 then you can't choose y as 5. Possible values of y are (10, 15, 20, 25). So, here four possible pairs of (x, y) possible { (5, 10), (5, 15), (5, 20), (5, 25)}.
Similarly, four possible pairs of (x, y) possible each time when x = 10, 15, 20 and 25.
$$ \therefore $$ Total number of ways in the first case = 5 $$ \times $$ 4 = 20.
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