Let's calculate what $a\times (b+c)$ actually is. By the distributive property,

$\qquad \qquad \qquad \qquad \qquad a\times (b+c) = a\times b + a\times c$

Substituting $a \times b + a \times c$ for $a \times (b+c)$ , we have the equation

$\qquad \qquad \qquad \qquad \qquad a \times b + a\times c = a\times b + c$

We can subtract $a \times b$ from both sides of the equation to obtain $a \times c = c$

Dividing $c$ from both sides, we have $a = 1$ . However, $c$ can also be equal to 0 (we can't divide by 0), since it ranges from 0 to 10 inclusive.

If $a=1$ , there are $11\times 11$ possibilities for $b$ and $c$ . Likewise, if $c=0$ , there are $11\times 11$ possibilities for $a$ and $b$ .

However, there are $11$ possibilities we must subtract, since there are $11$ possibilities for $b$ when $a=1$ and $c=0$

Hence, we have $121+121-11 = \boxed{231}\ \blacksquare$

By expanding the left-hand side, we get $ab+ac=ab+c$ Subtracting ab from each side gives us $ac=c$ Subtracting c from each side gives us $ac-c=0$ Taking out a factor of c gives us $c(a-1)=0$ Now we know that either $c=0$ or $a-1=0$ For each of these there are 11 options (each of the integers between 0 and 10 inclusive) for the other variable giving a total of 22. However, we must be careful to remove 1 instance of $a=1, c=0$ as this is duplicated. This gives us 21 different combinations for a and c. For each of the combinations, there are again 11 different possible values for b. Therefore the final answer is $21 \times 11= \boxed{231}$

Since, $a \times (b + c) = a \times b + c$ is required Therefore, $a \times c = c$ => $(a - 1) \times c = 0$ Hence, a = 1 or c = 0

• Case 1 : a = 1

Here, b can take 11 values and c can take 11 values => $1 \times 11 \times 11$ = 121 triplets.

• Case 2 : c = 0

In this case, a cannot be taken to be 1 because we have already considered that in previous case.

Hence , a can take 10 values and b can take 11 values => $10 \times 11 \times 1$ = 110 triplets.

Therefore, total triplets = 231

include <stdio.h>

#include <string.h>

main()
{
   int a,b,c,d,e, count;
   count=0;
   for(a=0;a<11; a++)
   {
       for(b=0;b<11; b++)
       {
           for(c=0;c<11; c++)
           {
               d=a*b+a*c;
               e=a*b+c;
               if(d==e)
                     count++;
            }
        }
    }
   printf("%d", count);
}

So the solution is $\boxed{231}$

Expanding, we get $ab+ac=ab+c$ so $ac=c$ Then, $b$ can be anything, giving 11 possibilities for $b$ .

If $c=0$ , then $a$ can be anything, giving 11 possibilities. If $c\neq 0$ , then $a=1$ , and $c$ can be anything except 0, giving 10 possibilities.

The answer is $11(11+10)=\boxed{231}$ .

We have: a(b+c) = ab +c

Expanding, ab + ac = ab + c

ac = c

This only occurs when a = 1, or c = 0.

If a = 1, b and c both range from 0 to 10 (11 possibilities each), giving us 11 x 11 = 121 ordered triples.

If c = 0, a and b both range from 0 to 10 (11 possibilities each), giving us 11 x 11 = 121 ordered triples.

Now, all that is left is to check which triples are repeated between the two sets. Obviously, the repeated triples will be the ones where a = 1 and c = 0. Hence, b ranges from 0 to 10 (11 possibilities).

The total number of ordered triples will be 121 + 121 - 11 = 231.

The equation is true when a=1 or c=0

for a=1 -- P(a), we have 11x11 = 121 possibilities (11 from b and 11 from c)

for c=0 -- P(c), we have 11x11 = 121 possibilities (11 from a and 11 from b)

for a=1 AND c=0 -- P(a and c), we have 11 possibilities

So for a=1 OR c=0 -- P(a or c) = P(a) + P(c) - P(a and c) = 121 + 121 -11 = 231

• $a(b+c) = ab + c \Rightarrow ab + ac = ab + c \Rightarrow ac = c \Rightarrow a = 1$ or $c = 0$ .
• There are 1 11 11 triples where $a = 1$ , and 11 11 1 triples where $c = 1$ .
• Note that if we simply sum up the two numbers above, we'll be counting instances where both $a = 1$ and $c = 0$ hold twice. There are 1 11 1 such instances, so we should subtract them once.
• $\Rightarrow 121+121-11 = \boxed{231}$ .

Distributing $a(b+c)=ab+c$ gives us $ab+ac=ab+c$ ,

or $ac=c$ ,

so a=1 or c=0

for all c=0 we have 121 ordered pairs from (0,0,0) to (0,10,0) (11 ordered pairs for each value of a, and we have 11 values from 0 to 10 inclusive)

for all a=1, c=/=0 we have another 110 ordered pairs from (1,0,1) to (1,10,1) (11 ordered pairs for each value of b, and we have 10 values from 1 to 10 inclusive)

Therefore our final answer is $121+110=\boxed{231}$

$a\times (b+c)=a\times b+ a\times c=a\times b+c \therefore a\times c=c$ If $c \neq 0 \Rightarrow a=1 \text{(I)}$ If $c=0 \Rightarrow a \text{ can be anything! (II)}$

(I) $1\times 11 \times 10 = 110$ (II) $11 \times 11 \times 1 = 121 \rightarrow 110+121=231$

Let's first consider these sets : $X = \{ 0 , 1 , 2 , 3 , \dots , 10 \} \ \ \ \ , \ \ \ \ Y = \{ 1 , 2 , 3 , \dots , 10 \}$ Now : $\because a \times (b+c) = a \times b + a \times c$ according to the right distributive property in multiplication , and $\because a \times ( b + c) = a \times b + c$ (given) . $\therefore a \times b + a \times c = a \times b + c$ Then by subtracting $(a \times b)$ from both sides (there is no problem in that) , we get : $\large a \times c = c$ Now let's assume that $c \neq 0$ that would be much easier , isn't it ?

Case 1 : ( $c \neq 0$ ) : $\ \ \therefore \ a = \frac{c}{c} = 1$ , in this case we have $a = 1 \ , \ b \in X$ but $c \in Y$ But what if $c = 0$ ?

Case 2 : ( $c = 0$ ) : By substituting $c$ we get : $a \times 0 = 0$ and this is satisfying , in this case we have $a , b \in X \ \ , \ c = 0$ .

Now in case 1 as we have $a=1$ fixed in the ordered triple $(a,b,c) = (1,b,c)$ so we look for the ordered pairs $(b,c)$ , so we get the product set of $X \times Y$ which contains $11 \times 10 = 110$ possible ordered pairs for $(b,c)$ and putting $1$ at the first of each ordered pair we have $110$ ordered triples .

In case 2 the same but we have $c=0$ fixed in the ordered triple $(a,b,c) = (a,b,0)$ so we look for the ordered pairs $(a,b)$ , so we get the product set of $X \times X$ which contains $11 \times 11 = 121$ possible ordered pairs for $(a,b)$ and putting $0$ at the end of each ordered pair we have $121$ ordered triples .

Summing all triples up $\Rightarrow \ 110 + 121 = \boxed{231}$ ordered triples .

We can simplify $a\cdot(b+c)=a\cdot b+c$ into $a\cdot b + a \cdot c=a\cdot b+c$ which is further simplified to $a \cdot c=c$ This will only be true if $a=1$ , or $c=0$ , or both. Let's count the cases:

$\text{Case} 1$ : Only $a=1$ . Then, for the ordered triple $a, b, c$ , we have $1$ choice for $a$ , (it can only be $1$ ), $11$ choices for $b$ , (it can be any integer between $0$ and $11$ ) and $10$ choices for $c$ (it can be any integer between $1$ and $10$ ; just not $0$ ). $1\cdot 11 \cdot 10=\boxed {110}$

$\text{Case} 2$ : Only $c=0$ . Then for the ordered triple $a, b, c$ , we have $10$ choices for $a$ , (it can be any integer between $0$ and $10$ ; just not $1$ ), $11$ choices for $b$ , (it can be any integer between $0$ and $11$ ) and $1$ choice for $c$ (it can be $0$ ). $10 \cdot 11 \cdot 1=\boxed {110}$

$\text{Case} 3$ : Both $a=1$ and $c=0$ . Then for the ordered triple $a, b, c$ , we have $1$ choice for $a$ , (it can only be $1$ ), $11$ choices for $b$ , (it can be any integer between $0$ and $11$ ) and $1$ choice for $c$ (it can be $0$ ). $1 \cdot 11 \cdot 1=\boxed {11}$

Adding these $3$ numbers together, we have $110+110+11=\boxed {231}$ different ordered triples of $(a, b, c$ that satisfy $a\cdot(b+c)=a\cdot b+c$ if $(a, b,$ and $c$ are all between $0$ and $10$ inclusive.

So expanding you have
$ab+ac=ab+c$

You can cancel $ab$ out to get

$ac=c$

From here you have 2 paths- either $a=1$ or $c=0$ and $a$ can be any value. After this it basically becomes a Combinatorics problem.

When $c=0$ you can $11*11=121$ ways of picking $a$ and $b$ since we have $0 \leq a,b,c \leq 10$

When $a=1, c \not=0$ you have 10 values of $c$ to choose from since it cant be 0 and 11 values of $b$ to choose from

Which gives $10*11=110$ ways of picking $c$ and $b$

So the total number of triples is

$121+ 110= \boxed{231}$

We can write a C++ Coding as follows;

include<constream.h>

void main()

{ clrscr();

int i,j,k,count=0,a,b;

for(i=0;i<=10;i++)

{ for(j=0;j<=10;j++)

{ for(k=0;k<=10;k++)

  {

   a = i*(j+k);

   b = i*j+k;

   if(a==b)

count++;

  }

}

}

cout<<"\n\n\tAns = "<<count;

getch();

return;

}

Output: Ans = 231

Lets write the two expressions after distributing $a$ .

Then we have:
$a \times b + a \times c == a \times b + c$

Obviously there are only two cases when the two equations are equal:

1. When $a == 1$

2. When $c == 0$

$a == 1$ $OR$ $c == 0$

When either of the two conditions are satisfied, both equations will be equal. Then for the 1st condition we have $11 \times 11$ possibilities, and for the 2nd one we also have $11\times 11$ possibilities. Now some triplets have been counted twice and those are when $c = 0$ $AND$ $a = 1$ . By inclusion and exclusion principle, we need to subtract these triplets from our answer, so the final answer is $(11 \times 11) + (11 \times 11) - (11) = \boxed{231}$

If we distribute a(b+c), we get ab+ac = ab + c. Then, we can subtract ab from both sides, ending with ac = c. Ordered pairs that satisfy this are ordered pairs with a = 1 or c = 0. Now we can do casework, counting the number of ordered pairs with (1,b,c) or (a,b,0). We have 11 possibilities for each letter, giving us 121 possibilities a = 1 and 121 for c = 1. However, we count any ordered pair (1,b,0) twice, which has 11 possibilities. Therefore, the grand total is 121+121-11=231.

If a x (b+c) = a x b + c, then a x c = c , then a = 1, * if c /= 0 * . We have: * if c /= 0 * the ordered triples are (1,b,c) with *c/= 0 * and so we have 11x10 = 110 triplets. If c = 0 we have axb = axb that is true for all a,b, so we have 11x11= 121 triplets like (a,b,0). In total we have 110+121=231 triplets.

Computer is always reliable for this kind of question. Answer: 231

we have two cases case 1. (when a=1) the possible number of triplets are 1 * 11 * 11=121 case 2. (when c=0) the possible number of triplets are ( 11 * 11 * 1)-(1 * 11 * 1)=110 we subtract because it has occurred in case 1. so the possible number of triplets which satisfy the property are 121+110=231

Did the same way !!!

counting using javascript

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var count=0;
for(var a=0;a<11;a++)
    for(var b=0;b<11;b++)
        for(var c=0;c<11;c++)
            if(a*(b+c)==a*b+c)
                count++;

console.log(count);

By expanding the equation, we get:

$a \times b + a \times c = a \times b + c$

$a \times c = c (1)$

Now, the original equation will be true when (1) is true.

Since b is eliminated completely from the result equation, b can take any value in range [0,10] without affecting the validity of (1). There are 11 possibilities for b.

For (1) to be true, either:

• $a = \frac{c}{c} = 1$ c can not be 0

a can only be 1. c can take any value in range (0,10], thus has 10 possibilities.

The number of combinations of a and c in this case is $1 \times 10 = 10$

• $c = 0$

a can take any value in range [0,10], thus has 11 possibilities. c can only be 0.

The number of combinations of a and c this time is $1 \times 11 = 11$

The number of total possible combination of a and c is 11 + 10 = 21.

The number of combinations of a,b and c are $11 \times 21 = 231$

that equation is correct if a=1, or c=0

number of orders with (a=1) = 11*11=121, as we have 11 numbers from 0 to 11 could be the values of b and c

number of orders with (c=0) = 11*11=121, as we have 11 numbers from 0 to 11 could be the values of a and b

number of orders with (a=1) and (c=0) = 11, as we have 11 numbers from 0 to 11 could be the values of b

number of orders with (a=1) or (c=0) =number of orders with (a=1)+number of orders with (c=0)-number of orders with (a=1) and (c=0)=121+121-11= $\boxed{231}$

a(c-1)=0

Possibilities:

if a=1 then c can be from 0 to 10. if c=1 then a can be from 1 to 10. so totally 21 possibilities. b can be from 0 to 10, so 11 posibilities. 21*11=231

(axb)+(axc)=(axb)+c =>ac=c...this means it does not depend on b....so every 11 values of b is possible for each duplet (a,c)...so a factor of 11 is included to the number of triplets... now, for c not equal to 0,a=1 therefore, all duplets (a,c)= (1,1),(1,2)...(1,10)...that is 10 duplets are possible.... again,if c=0,all 11 values of a is valid...therefore 11 duplets so total number is=11x(11+10)=231

When c = 0, a and b both can take 11 values (0..10 inclusive) that gives us 11 x 11 possible triplets. When c = (1..10), a has to be 1 and b can still take 11 values. That gives us another 11 x 10 possible triplets. Add 121 + 110. That is 231.