Counting Ordered Pairs

one of them has to be 5 the other can be between 1 an 10 so 10 ways. You can switch x and ys around so double it to get 20 But you just counted 5,5 twice so take away 1 to get:

$\boxed{19}$

The equation $(x-5)(y-5)=0$ is only correct if x=5 or, y=5 or, both.

Now, when x=5, for every y with 1<=y<=10 we get an ordered pair which is a solution.

For example, (5,1),(5,2),(5,3).....(5,9),(5,10) every ordered pair satisfies (x-5)(y-5)=0

Note, (5,5) is a possible ordered pair.

So, there are 10 ordered pairs.

Now compute for y=5.

for every x with 1<=x<=10 we get ordered pairs (1,5),(2,5),.....(10,5) which satisfies our problem.

Note, We again get a (5,5) in this case.

Now note that, (1,5) and (5,1) are two different ordered pair. But, (5,5) and (5,5) is same.

So, we have totally 10+10-1(as there is an ordered pair which is common to both cases)=19 different ordered pairs.

The eqn holds true i) for x=5,y=any no. and ii)for y=5,x=any no. As we have range from 1 to 10, for i) we have 10 solutions, for ii) we have 10 solutions. Now adding 10 and 10 and subtracting a common solution (5,5), we get ans. 19

//include iostream statement comes here. Rest follows

using namespace std;

bool isPair (const int& x, const int& y) {

if ((x - 5) * (y - 5) == 0) {

return (true);

}

else {

return (false);

}

}

int main (int argc, char* argv []) {

int counter (0);

for (int x = 1; x <= 10; x++) {

for (int y = 1; y <= 10; y++) {

  if (isPair (x, y) == true) { counter++; }

}

}

cout << counter << endl;

return (0);

}

with x=5 there are 10 possible values of y with y=5 there are 10 ppossible values of x .answer is 20 but there are two ordered pairs with (5,5)therefore answer if 19

if you want (x-5)(y-5)=0 then x 0r y must be 5. and as x and y both are >=1 and <=10 we have 10 pair for each x=5 and y=5. So total 20 pair. But in this 20 pair (5,5) is repeating so 19 pairs.

We have two cases:

$x-5 = 0$

We have 1 choice for x and 10 for y, then we have 10 ordered pairs

$y-5 = 0$

We have 10 choices for x and 1 for y, so we have 10 ordered pairs

Thus, we have 20 ordered pairs, but we must take one, because a pair was counted twice (the pair $(5.5)$ )

So are 19 ordered pairs.

If the first factor is zero then x will be 5 and y may be any no from 1to10 and hence the ordered pairs are (5,1);(5,2);(5,3);(5,4);(5,5);(5,6);(5,7);(5,8);(5,9) and (5,10).

Similarly if the second factor is zero then the value of y will be 5 and the value of x will be any no from 1 to 10 and hence the ordered pairs are (1,5);(2,5);(3,5);(4,5);(4,5);(5,6);(5,7);(5,8);(5,9) and (5,10) as we can not take (5,5) two times.

we have to give 5 value for x or y or both
for x=5 no of values of y is 10 then no of ordered pairs are 10
similarly for y=5 no of ordered pairs 10
in those ordered pairs (5,5) repeated
total pairs 10+10-1=19

For x = 5 or y = 5, we find 19 unique solutions: {(5,1),(5,2),...,(5,5),...,(5,10),(1,5),(2,5),...,(10,5)} Of course you may not count (5,5) twice!

As long as either x or y equals 5, then the result will be zero. So if x is 5, then there will be 10 other numbers to choose y. If y is 5, there will be 10 ways to choose x. So it would seem that the answer would be 20, but it's actually 19, because it counts x=5, y=5 twice.

We first notice that either $x=5$ or $y=5$ . Start by assuming that $x=5$ . There are 10 such possibilities as $y$ can independently take on 10 values. Next, consider the case where $y=5$ . In this case there are also 10 possibilities, but we have counted one twice, the one where both $x$ and $y$ are equal to $0$ . So the total amount of ordered pairs is $10+10-1=\boxed{19}$

X=5 then y will any of 1 to 10

when y=5 then x=1to 10

but there is a common pair (5,5)

so we have 10+10-1=19 ordered pairs

Both x=5 and y=5 solve the equation as well as any combination containing at least one of the two. But, since we are looking for ordered pairs, (5,5) can only be counted once.

i.e.:

(5,1), ....., (5,5), .......(5,10) & (1,5), ....., (5,5), ........(10,5) ↪there are 19 ordered pairs.

• Para x = 5 e y vai de 1 à 10: --------> (5,1),(5,2)(5,3),(5,4)(5,5)....(5,10).

• Para y = 5 e x varia de 1à 10: -------> (1,5),(2,5)(3,5),(4,5)(5,5)....(10,5).

O total é 20, mas como temos o par ordenado (5,5) repetido subtraímos 1.

Assim teremos -------> 20 -1 = $\boxed{19}$ pares

(x-5)(y-5)=0 if either (x-5) or (y-5) or both are zero. This happens when either x or 5 or both are equal to 5. The possible ordered pairs are 9 of the form (x,5) , 9 (5,y) and (5,5). That gives a total of 19 ordered pairs

If we want the equation is equal to 0 we need one part of the bracket is 0.

fo example:(x-5)(5-5)=0 or (5-5)(y-5)=0

hence (9)(1)+(9)(1)+1=19

(x−5)(y−5)=0 <=> x = 5 or y = 5. When x = 5, y can be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 <=> we have 10 pair of (x, y). When y = 5, x can be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 <=> we have 10 pair of (x, y). But in 20 pair of (x, y) have two same pair (5, 5). So, the result is 19