Absolute integer sequences

Let the sum of the positive integers in $S$ be $p$ , and let the sum of the negative integers be $q$ . We know that $p+q=0$ $p-q=4$ Therefore, $p=2, q=-2$ . Now, positive integers that add to 2 are either $2$ or $1,1$ , and similarly, negative integers that add to -2 are either $-2$ or $-1,-1$ . This gives us the following four possibilities: $(-1,-1,1,1,0,0,0,0)$ $(-1,-1,2,0,0,0,0,0)$ $(-2,1,1,0,0,0,0,0)$ $(-2,2,0,0,0,0,0,0)$ Each of these can be permuted, since $S$ is order-dependent. This gives us the answer $\dfrac{8!}{2!2!4!}+\dfrac{8!}{2!5!}+\dfrac{8!}{2!5!}+\dfrac{8!}{6!}=420+168+168+56=\boxed{812}$

Take the second condition, that $\displaystyle\sum_{i =1}^8 \left|s_i\right| = 4.$ Since $\left|s_i\right|$ is a nonnegative integer, there are a limited number of possible tuples $\left( \left|s_1\right|, \left|s_2\right|, \ldots, \left|s_8\right|\right).$ Taken without regard to order, this tuple must be one of the following:

• $(4, 0, 0, 0, 0, 0, 0, 0)$

• $(3, 1, 0, 0, 0, 0, 0, 0)$

• $(2, 2, 0, 0, 0, 0, 0, 0)$

• $(1, 1, 2, 0, 0, 0, 0, 0)$

• $(1, 1, 1, 1, 0, 0, 0, 0)$

Now we need to consider the first condition for each tuple: we must see if it is possible to change the signs of some of the numbers in each tuple to produce a sum of $0.$

• It's impossible for $(4, 0, 0, 0, 0, 0, 0, 0),$ because the only possible sums are $\pm 4.$

• It's impossible for $(3, 1, 0, 0, 0, 0, 0, 0),$ because the only possible sums are $\pm 4,$ $\pm 2.$

• It is possible for $(2, 2, 0, 0, 0, 0, 0, 0),$ if the sequence is a permutation of $(2, -2, 0, 0, 0, 0, 0, 0).$

• It is possible for $(1, 1, 2, 0, 0, 0, 0, 0),$ if the sequence is a permutation of $(1, -1, 2, 0, 0, 0, 0, 0).$

• It is possible for $(1, 1, 1, 1, 0, 0, 0, 0),$ if the sequence is a permutation of $(1, 1, -1, -1, 0, 0, 0, 0).$

Therefore, the sequence $(s_1, s_2, \ldots, s_8)$ must be a permutation of one of the following:

$\begin{aligned} &(2, -2, 0, 0, 0, 0, 0, 0) \\ &(1, -1, 2, 0, 0, 0, 0, 0) \\ &(1, 1, -1, -1, 0, 0, 0, 0) \end{aligned}$

There are $8 \cdot 7 = 56$ ways to order the first sequence, $8 \cdot 7 \cdot 6 = 336$ ways to order the second one, and $\binom{8}{2} \binom{6}{2} = 420$ ways to order the third one. Therefore, there are $56 + 336 + 420 = \boxed{812}$ such sequences.

There are only three types of possibilities in the sense described below in which $s_i$ are non-zero.

1. Four of them non-zero. In this case each non-zero $|s_i| = 1$ . Then two of the are +1 and the other two are -1. The number of possibilities are ${8 \choose 4}\times {4 \choose 2} = 420$ The first factor is for choosing 4 from 8, the second factor is for choosing two 1's from 4 chosen numbers.

2. Three of them are non-zero. In this case two of them are such that $|s_i| = 1$ and the third is $|s_i| = 2$ . The number of possibilites are ${8 \choose 3}\times {3 \choose 2} \times 2 = 336$ The first factor is for choosing 3 from 8, the second factor is for choose two 1's or $-1$ 's and the third factor is for sign revesal.

3. Two of them are non-zero. In this case one of them is 2 and the other is $-2$ . Given a section of 2 of 8, there are two ways. Thus the number of possibilites are ${8 \choose 2}\times 2 = 56$

Summing them gives the answer.

$\sum_{i=1}^{8} |S_i| = 4 \Rightarrow 4 \leq n \leq 6,$ where $n$ is the number of $0's$

Case 1: $n = 6,$ so $S_1=S_2=S_3=S_4=S_5=S_6= 0$

$S_7 =-S_8$ and $|S_7| + |S_8| = 4$ so they are $2,-2$

Case 2: $n = 5,$ so $S_1=S_2=S_3=S_4=S_5 = 0$

$S_6+S_7+S_8 = 0$ and $|S_6|+|S_7|+|S_8| = 4$ so they are $1,1,-2$ or $-1,-1,2$

Case 3: $n=4.$ so $S_1=S_2=S_3=S_4 =0$

$S_5+S_6+S_7+S_8 = 0$ and $|S_5|+|S_6|+|S_7|+|S_8| = 4$

So they are $1,1,-1,-1$ in some order.

Total number of possible sequences = $\frac{8!}{4!*2!*2!} + \frac{8!*2}{5!*2!} + \frac{8!}{6!}$

= $420 + 336 + 56 = 812$

To satisfy the second condition, we have:

(1) - Or 4 of the numbers are +/-1, and the other 4 numbers are 0. (2) - Or 2 of the numbers are +/-1 and 1 of them is +/-2, and the other 5 numbers are 0. (3) - Or 1 of them is +/1 and 1 of them is +/-3, and the other 6 numbers are 0. (4) - Or 2 of the numbers are +/- 2, and the other 6 numbers are 0. (5) - Or 1 of the numbers is +/- 4, and the other 7 numbers are 0.

Well, to satisfy the first condition, we can use, of the numbers above:

(1) - If two of the numbers are 1, two are -1 and 4 are 0 (2) - If two of the numbers are 1, one is -2 and 5 are 0 (2") - If two of the numbers are -1, one is 2 and 5 are 0 (4) - If one of the numbers is 2, one is -2 and 6 are 0

Now, we need to calculate the permutations of each of the possibilities. So, (1) - $8!/4!.2!.2!$ = 420 (2) - $8!/5!.2!$ = 118 (2") - $8!/5!.2!$ = 118 (4) - (8!/6!) = 56

Adding them, you find 812, the solution of the problem.

From given, sum of 8 numbers is 0 and sum of their absolute values is 4. => that each integer of S is less than 4. possibilities for a sum of 4 are as follows: 1,1,1,1 -- OK(can be 2 1's and 2 -1's) so possible sequences for this are {8 \choose 4} * {4 \choose 2} 1,1,2 -- OK(can be 2,-1,-1 or -2,1,1) so possible sequences for this are {8 \choose 3} * {3 \choose 1} * 2 2,2 -- OK(can be 2,-2) so possible sequences for this are {8 \choose 2} * {2 \choose 1} 3,1 -- NOT OK

Hence final answer is sum of all the possible number of sequences as mentioned above which is 812

From the condition of $\sum_{i=1}^8 |s_i| = 4$ , we can narrow down the selection to $(1, 1, -1, -1, 0, 0, 0, 0),$ $(1, 1, -2, 0, 0, 0, 0, 0),$ $(-1, -1, 2, 0, 0, 0, 0, 0),$ $(2, -2, 0, 0, 0, 0, 0, 0)$

For $(1, 1, -1, -1, 0, 0, 0, 0),$ there are $\frac{8!}{4! \times 2! \times 2!} = 420$ permutations. For $(1, 1, -2, 0, 0, 0, 0, 0),$ there are $\frac{8!}{5! \times 2!} = 168$ permutations. For $(-1, -1, 2, 0, 0, 0, 0, 0),$ there are $\frac{8!}{5! \times 2!} = 168$ permutations. For $(2, -2, 0, 0, 0, 0, 0, 0),$ there are $\frac{8!}{6!} = 56$ permutations.

Adding all these up, the total number of sequences is $812$ .

Existem 3 possibilidades de algarismo para atender as premissas:

1) Aparecerem 4 zeros, dois "-1" e dois "1"

2) Aparecerem 6 zeros, um -"2" e um "2"

3) Aparecerem 5 zeros, ou (dois "-1" e um "2") ou (dois "1" e um "-2")

Com isso, basta encontrar o número de permutações em cada caso:

1) 8!/(4! 2! 2!) = 420

2) 8!/6! = 56

3) 8!/(5!*2!) x 2 = 336 (Obs.: multipliquei por 2 para contar a troca de sinal entre os "1" com o "2")

Assim: 1 + 2 + 3 = 812

We can see that $|s_i| \le 2$ for all $i$ , since if there is a $|s_i| > 2$ , then $\sum_{i = 1}^8 s_i > 4$ , a contradiction. Now, if there are two nonzero elements $s_i$ , the two nonzero elements must be $-2$ and $2$ , giving a total of $\binom{8}{2} = 28$ sets. If there are three nonzero elements $s_i$ , one nonzero element must be $\pm 2$ and the other two $\mp 1$ , which gives $2\binom{8}{1}\binom{8}{2} = 336$ sets. If there are four nonzero elements $s_i$ , the four elements must be $\pm1$ , yielding $\binom{8}{2}\binom{6}{2} = 420$ sets. There cannot be zero or more than four nonzero elements, or $\sum_{i = 1}^8|s_i| > 4$ , a contradiction. The total number of sets is then $28 + 336 + 420 = \fbox{812}$ sets.

S={ 2,-2,0,0,0,0,0,0 } could be a possible sequence but we can arrange integer of this sequence in 8!/6!=56 way,so we get 56 sequence

similarly,for S={1,1,-1,-1,0,0,0,0} we get 8!/(4! 2! 2!)=420 sequence;
for S={2,-1,-1,0,0,0,0,0} we get 8!/(5! 2!)=168 sequence;
for S={-2,1,1,0,0,0,0,0} we get 8!/(5!
2!)=168 sequence;

any sequence other than above mentioned sequence is not possible.

so,total no of sequence=56+420+168+168=812 (Ans).