Negation of a double negative is not ... Mind Blown

$1+1+1+1+1+1+1+1+1+1+1+1=12$

So to get $4$ we have to decrease it by $8$ . To do that we got to have $4$ $-1$ 's.

There are $11$ $+$ signs and we need $4$ $-$ signs.

We have $11C4=330$ .

We know that there are: 12 "1"s and 11 "+" signs.

To interpret the question, "How many ways can a subset of the + signs..." . They are asking How many ways can a number (<12) of + signs be turned into - signs so that the eqn becomes valid

To fulfill the equation we must remove 8 "1", meaning that we must change 4 "+" signs into $\boxed{4}$ "-" signs.

There are no brackets in this question, so we don't have to worry about that.

Now, we know that we must choose any 4 of the 11 "+" to change, thus, 11C4 = 330

by permutations and combinations: we have to replace any of the 11 "+" signs with 4 "-"signs to get the ans ;therefore 11C4=330 Q.E.D

4 can be obtained by turning 4 + signs into - signs.

For this we have to take a combination of 4 + signs out of 11. This can be done in 11C4 ways i.e. 330 ways.

Total No. of Signs=11, We observe that 8-4=4, so there must be 4 negative signs out of 11 signs. So, Number of ways is simply 11C4=330

Para que a equação se torne válida, é necessário que tenham apenas 4 sinais +. O restante, há de se anularem.

(+)1 +1 +1 +1 +1 -1 +1 -1 +1 -1 +1 -1

Cai em uma permutação com repetição. 11! / (7!.4!) = 330

Well, I did it like this: Firstly, there are 12 1's of which, if there are 4 -1's and 8 1's we get the desired result. Also, there are 11 +signs and the first '1' is positive so we need to change any 4 out of the 11 '+' signs to '-'. So, definitely the answer is 11C4 = 330.

The sum given is $12$ in Left hand side of the expression .We need to alter some $'+'$ into $'-'$ to make it $4$ . We observe, $8+4=12$ and $8-4=4$ .Therefore, we need to alter $4$ $'+'$ signs into $'-'$ signs out of a total of $11$ $'+'$ signs.This can be done in: $\dbinom{11}{4}=\boxed{330}$ ways.

There are 12 of the number 1.Each time we flip a sign from $+$ to $-$ we subtract 2 because before the flip the 1 was a positive 1 and after the flip it is a negative 1 so the difference is $1 -(-1) = 2$ .So we must flip $\frac{12-4}{2} = 4$ signs.The number of ways we can pick the signs to be flipped is $11 C 4 = \frac{11 \times 10 \times 9 \times 8}{ 4 \times 3 \times 2 \times 1} = 330$ (the combination formula).

11!/4!7!

Note that the only way to have our expression evaluate to 4 is to have 4 negative signs and 7 positive. This should be clear since: $(7+1)+(-1)(4)=4$ (we have the $7+1$ term as the first digit is positive).

So, out of $11$ total symbols, we must choose $4$ negative signs, and, since all negative signs are equivalent, it is order-independent.

We compute this by: $\binom{11}{4}=330$ Which is our answer.

You need only 4 1's with + to make 4 means turn remaining 7 '+' to '-' . It turns into a problem of making permuting 11 objects of which 7 are like of one kind and 4 are like of second kind. 11!/4!*7!=330