Let x and y be distinct integers where 1 ≤ x ≤ 25 and 1 ≤ y ≤ 25 . Then, the number of ways of choosing x and y , such that x + y is divisible by 5 , is _____ .

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Given: 1 ≤ x, y ≤ 25 And, x, y are distinct integers. Let x + y = 5k , where k ∈ N So, we have x y Number of ways 5λ i.e, 5, 10, 15, 20, 25 5λ i.e, 5, 10, 15, 20, 25 Since, x and y are distinct integers, so we cannot pair (5, 5), (10, 10), . . . (25, 25) 20 5λ + 1 i.e., 1, 6, 11, 16, 21 5λ + 4 i.e., 4, 9, 14, 19, 14 25 5λ + 2 i.e, 2, 7, 12, 17, 22 5λ + 3 i.e., 3, 8, 13, 18, 23 25 5λ + 3 i.e., 3, 8, 13, 18, 23 5λ + 2 i.e, 2, 7, 12, 17, 22 25 5λ + 4 i.e., 4, 9, 14, 19, 14 5λ + 1 i.e, 1, 6, 11, 16, 21 25 Total number of ways = 120

$x$	$y$		Number of ways
$5\lambda$ i.e, $5, 10, 15, 20, 25$	$5\lambda$ i.e, $5, 10, 15, 20, 25$	Since, $x and y$ are distinct integers, so we cannot pair \left(5, 5\right), \left(10, 10\right), . . . \left(25, 25\right)	$20$
$5\lambda + 1$ i.e., $1, 6, 11, 16, 21$	$5\lambda + 4$ i.e., $4, 9, 14, 19, 14$		$25$
$5\lambda + 2$ i.e, $2, 7, 12, 17, 22$	$5\lambda + 3$ i.e., $3, 8, 13, 18, 23$		$25$
$5\lambda + 3$ i.e., $3, 8, 13, 18, 23$	$5\lambda + 2$ i.e, $2, 7, 12, 17, 22$		$25$
$5\lambda + 4$ i.e., $4, 9, 14, 19, 14$	$5\lambda + 1$ i.e, $1, 6, 11, 16, 21$		$25$

Let x and y be distinct integers where 1 ≤ x ≤ 25 and 1 ≤ y ≤ 25 ....

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