I'm not sure where to start on this question. If I label the four digits $a,b,c,d$, what I did was split the four digits in two and find the number of ways $a+b=1$ and the number of ways $c+d=9$, and then $a+b=2$ and $c+d=8$ and so on until $a+b=5$ and $c+d=5$. What is a way I could solve a problem such as this logically so that I could approach similar problems in a similar way?
Asked
Active
Viewed 312 times
3 Answers
2
There is no set formula. You will have to find the best possible way to split the problem into cases. For this specific question as the sum is just $10$, below is one of the ways.
Please note that $~1 + 2 + 3 + 4 = 10$
Preserving the order of the digits -
If we make the first digit $0$, $1$ gets added to the last digit. If we take out $1$ each from first two digits, either $1$ gets added to third and fourth or $2$ gets added to the last. If we take out $1$ each from first three digits, it must get added to the last digit.
That is $5$ solutions in total. If the order of the digits matters then the number of solutions is $5 \cdot 4!~$.
Math Lover
- 51,819
0
First notice the numbers that would not work in any combination with the other numbers(8 and 9) leaving 0-7.
If you enumerate all cases you can quickly find the number of combinations:
Example: For 7 the only combination that works is 0,1,2,7 The only combination containing 6 is 0,1,3,6 The only combinations containing 5 are 0,1,4,5 and 0,2,3,5 This leaves the combination that hasn't been used yet: 1,2,3,4
You may already understand this intuitively but if you check 3 there will be no combination that has not already been counted previously(and so on for 2,1,0).
iceninja21
- 111
0
Alternative approach:
The approach used by the OP (i.e. original poster) seems best for this problem, because the numbers were so small. If instead, you had (for example) been asked to have the sum equal $1000,~$then your method would not have generalized well.
For a more general method:
Suppose that you are required to enumerate how many solutions there are to :
$x_1 + x_2 + x_3 + x_4 = n ~: ~n~$ is some fixed (known) large positive integer.
$x_1, x_2, x_3, x_4 \in \Bbb{Z_{\geq 0}}.$
$x_1 < x_2 < x_3 < x_4.$
There are a couple of ways to go about this. First, I will present the method that I recommend here, specifically because there are $4$ or fewer variables. Then, I will present a different (uglier) method that would probably be the better approach if there were more than $4$ variables.
Suppose that you interpret this as a Stars and Bars problem, as discussed here and here.
As a first step, you would enumerate the number of non-negative integer solutions to $x_1 + x_2 + x_3 + x_4 = n.$ This approach would then require the following two adjustments:
First, you would have to enumerate and deduct all of the possible solutions where at least $2$ of the $4$ variables are equal. This is fairly routine, when working with only $3$ variables. For $4$ variables, it is somewhat more difficult, but is still better than the ugly method that I will show at the end of my response.
Then, after the deduction, you would have identified all possible solutions, where the $4$ variables are distinct. You would then have to divide this resulting enumeration by $(4!)$, since you want the variables to be in strictly ascending order.
For illustrative purposes, I am going to specifically enumerate the number of satisfying solutions to
$$x_1 + x_2 + x_3 + x_4 = 1000. \tag1 $$
I am going to employ Inclusion-Exclusion. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.
Let $S$ denote the set of all non-negative integer solutions to the equation in (1) above, without any regard for whether the variables are equal or not, or whether the variables are in ascending order.
Let $S_1$ denote the subset of $S$ where $x_1 = x_2$.
Let $S_2$ denote the subset of $S$ where $x_1 = x_3$.
Let $S_3$ denote the subset of $S$ where $x_1 = x_4$.
Let $S_4$ denote the subset of $S$ where $x_2 = x_3$.
Let $S_5$ denote the subset of $S$ where $x_2 = x_4$.
Let $S_6$ denote the subset of $S$ where $x_3 = x_4$.
For any set $E$ with a finite number of elements, let $|E|$ denote the number of elements in $E$.
Then, as discussed, the desired enumeration will be
$$\frac{|S| - |S_1 \cup S_2 \cup \cdots \cup S_6|}{4!}. \tag2 $$
Let $T_0$ denote $|S|.$
Let $T_1$ denote $~\displaystyle \sum_{k=1}^6 |S_k|.$
For $r \in \{2,3,4,5,6\}$, let $T_r$ denote $~\displaystyle \sum_{1 \leq i_1 < i_2 < \cdots < i_r \leq 6} |S_{i_1} \cap S_{i_2} \cap \cdots \cap S_{i_r}|.$
That is, $T_r$ denotes the sum of $~\displaystyle \binom{6}{r}~$ terms.
Then, in accordance with Inclusion-Exclusion, the enumeration represented in (2) above will be
$$\frac{\sum_{r=0}^6 (-1)^r T_r}{4!}. \tag3 $$
So, to complete the problem represented in (1) above, it only remains to compute each of $T_0, T_1, T_2, T_3, T_4, T_5,$ and $T_6$.
$\underline{\text{Computation of} ~T_0}$
This is a standardard Stars and Bars enumeration.
$$T_0 = \binom{1000 + 3}{3} = \binom{1003}{3} = 167,668,501.$$
$\underline{\text{Computation of} ~T_1}$
$|S_1|$ represents the number of solutions where $x_1 = x_2$. Here, $x_1$ can run from $0$ through $500$. For each such value, with $x_2 = x_1$, I will then need to enumerate the number of solutions to
$x_3 + x_4 = 1000 - 2x_1 ~: ~(1001 - 2x_1)~$ solutions.
So, $~\displaystyle |S_1| = \sum_{k=0}^{500} (1001 - 2k).$
This equals $~\displaystyle (1001 \times 501) - 2\left[\frac{500 \times 501}{2}\right] = 251,001.$
By reasons of symmetry, you have that
$|S_1| = |S_2| = \cdots = |S_6|.$
Therefore,
$$T_1 = 6 \times 251,001 = 1,506,006.$$
$\underline{\text{Computation of} ~T_2}$
$|S_1 \cap S_2|$ represents the number of solutions where $x_1 = x_2$ and $x_1 = x_3$. Here, with $x_1 = x_2 = x_3,$ $x_1$ can run from $0$ through $333$.
Therefore, $~|S_1 \cap S_2| = (334).$
By reasons of symmetry, you have that of the $~\displaystyle \binom{6}{2} = 15$ different ways of combining $2$ of the $6$ subsets, all but $3$ of them will have the same enumeration of $(334)$.
The exceptions are $~|S_1 \cap S_6|, |S_2 \cap S_5|,~$ and $~|S_3 \cap S_4|.$
$|S_1 \cap S_6|$ represents the number of solutions where $x_1 = x_2$ and $x_3 = x_4$. Here, you can have $x_1$ run from $0$ through $500$, so there are $501$ solutions.
Therefore,
$$T_2 = (12 \times 334) + (3 \times 501) = 5511. $$
$\underline{\text{Computation of} ~T_3}$
Analysis here is very similar to that of the computation of $T_2$.
The $~\displaystyle \binom{6}{3} = 20$ ways of combining $3$ of the $6$ subsets will have two patterns.
$4$ of the $20$ ways are represented by $|S_1 \cap S_2 \cap S_4|$, which represents having $x_1 = x_2 = x_3,$ which has an enumeratiom of $334$.
The other $16$ ways all result in $x_1 = x_2 = x_3 = x_4$. Because $n = 1000$ happens to be a multiple of $4$, each of these $16$ subset unions has $1$ solution rather than $0$ solutions.
So,
$$T_3 = (4 \times 334) + (16 \times 1) = 1352.$$
$\underline{\text{Computation of} ~T_4}$
Analysis here is very similar to that of the computation of $T_3$.
The $~\displaystyle \binom{6}{4} = 15$ ways of combining $4$ of the $6$ subsets will all require that $x_1 = x_2 = x_3 = x_4$.
Therefore,
$$T_4 = 15.$$
$\underline{\text{Computation of} ~T_5}$
Analysis here is very similar to that of the computation of $T_4$.
The $~\displaystyle \binom{6}{5} = 6$ ways of combining $5$ of the $6$ subsets will all require that $x_1 = x_2 = x_3 = x_4$.
Therefore,
$$T_5 = 6.$$
$\underline{\text{Computation of} ~T_6}$
Analysis here is very similar to that of the computation of $T_6$.
The $~\displaystyle \binom{6}{6} = 1$ way of combining all $6$ of the $6$ subsets will all require that $x_1 = x_2 = x_3 = x_4$.
Therefore,
$$T_6 = 1.$$
Final enumeration
$$\frac{\sum_{r=0}^6 (-1)^r T_r}{4!} $$
$$= \frac{(167,668,501 + 5,511 + 15 + 1) - (1,506,006 + 1,352 + 6)}{24}$$ $$= \frac{166,166,664}{24} = 6,923,611. \tag4 $$
The second approach to identifying the solutions in (1) above is as follows:
Let $y_1 = x_2 - x_1 \implies x_2 = x_1 + y_1$.
Let $y_2 = x_3 - x_2 \implies x_3 = x_1 + y_1 + y_2$.
Let $y_3 = x_4 - x_3 \implies x_4 = x_1 + y_1 + y_2 + y_3$.
Then, the following constraints must be satisfied:
- $x_1 + x_2 + x_3 + x_4 = 1000$.
- $x_1, x_2, x_3, x_4$ are all non-negative integers.
- $y_1, y_2, y_3$ must all be positive integers.
Therefore, you must satisfy:
- $4x_1 + 3y_1 + 2y_2 + y_3 = 1000$
- Where $x_1 \in \Bbb{Z_{\geq 0}}$
- And $y_1, y_2, y_3 \in \Bbb{Z^+}.$
The trick here is to work backwards.
For any real number $r$, let $\lfloor r\rfloor$ (i.e. the floor function) denote the largest integer $\leq r.$
Let $f(t)$ denote the number of solutions to $2y_2 + y_3 = t.$ Since $y_2$ and $y_3$ must both be positive integers, $y_2$ will run from $1$ through $~\displaystyle \left\lfloor \frac{t-1}{2} \right\rfloor.$
Therefore,
$$f(t) = \left\lfloor \frac{t-1}{2} \right\rfloor.$$
Let $g(t)$ denote the number of solutions to $3y_3 + 2y_2 + y_3 = t.$ Since $y_1,y_2,$ and $y_3$ must all be positive integers, $y_3$ will run from $1$ through $~\displaystyle \left\lfloor \frac{t-3}{3} \right\rfloor.$
Therefore,
$$g(t) = \sum_{y_1 = 1}^{\left\lfloor \frac{t-3}{3} \right\rfloor} f(t - 3y_1) = \sum_{y_1 = 1}^{\left\lfloor \frac{t-3}{3} \right\rfloor} \left\lfloor \frac{t-1 - 3y_1}{2} \right\rfloor .$$
The desired overall computation is the number of solutions to $4x_1 + 3y_3 + 2y_2 + y_3 = 1000.$ Since $y_1,y_2,$ and $y_3$ must all be positive integers, $x_1$ will run from $0$ through $248.$
For each such value of $x_1$, the number of corresponding solutions will be $g(1000 - 4x_1).$
Therefore, the overall computation is
$$\sum_{x_1 = 0}^{248} \left[ \sum_{y_1 = 1}^{\left\lfloor \frac{1000 - 4x_1 - 3}{3} \right\rfloor} \left\lfloor \frac{1000 - 4x_1 - 1 - 3y_1}{2} \right\rfloor \right]. \tag5 $$
This is an opportune moment to discuss computer-sanity-checking, which refers to verifying your analysis by writing a computer program to brute force loop through all of the possible pertinent values of the variables, counting the number of combinations that result in a satisfying solution.
Unfortunately, most PC's will balk at greater than $(10)^8$ cases to loop through. If I had been interested in computer-sanity-checking the computation in (4) above, I would have written a program in Java to enumerate the analogous problem of $x_1 + x_2 + x_3 + x_4 = 100$. Then, I would have manually repeated the analysis/computations that I posted, applying them to the altered $n = 100$ equation, to verify my methods.
If you have never programmed before, I advise making Python your first programming language.
As another wrinkle, I normally would not bother trying to compute the value in $(5)$ explicitly. However, here, you've got two different approaches to the same problem, with no (other) sanity-checking. Therefore, I will method-sanity-check by semi-brute-force doing the computations in (5) above, to verify that the result matches the result in (4) above.
Edit
Actually, although my overall analysis was okay, initially, the results from the two methods did not match. I had made some arithmetic mistakes, that (for some reason), going back to proofread didn't catch.
I ended up computer-sanity-checking both methods, which allowed me to identify the location where I transposed the digits of a number. I mention this, because I was able to creatively have the computer program loop through the appropriate values, exploring both strategies, even though the upper bound was $n = 1000$.
For the first approach :
* This method creates a 3 fold inner loop.
* I1 goes from 0 thru 249.
* I2 goes from I1 + 1 thru 333.
* I3 goes from I2 + 1 thru 500.
* I4 = 1000 - I1 - I2 - I3.
* If I4 > I3, then Ctr + 1.
When computer-sanity-checking the 2nd approach, (pseudocoded below), I checked each congruence class, (mod 3) separately.
* Loop intI1 from 1 thru 83.
* IntX1 = (3 * intI1) - k, k in {1,2,3}.
* Loop IntY1 from 1 thru ((997 - (4 * intX1))/3).
* intCtr = intCtr + (997 - ((4 * intX1) + (3 * intY1))) /2.
$\underline{\text{Relevant formulas}}$
$\displaystyle \sum_{k=1}^n (k) = \left[\frac{n(n+1)}{2}\right].$
$\displaystyle \sum_{k=1}^n (k)^2 = \left[\frac{n(n+1)(2n+1)}{6}\right].$
$\displaystyle \sum_{k=1}^n (T + rk) = nT + k\left[\frac{n(n+1)}{2}\right].$
Let $A$ denote the set $\{0,3,6,9,\cdots,246\}.$
Let $B$ denote the set $\{1,4,7,10,\cdots,247\}.$
Let $C$ denote the set $\{2,5,8,11,\cdots,248\}.$
As $x_1$ goes from $0$ through $248,$ it will take on each of the values in each of the sets $A,B,$ and $C$.
For $~\displaystyle x_1 \in A, ~\left\lfloor \frac{1000 - 4x_1 - 3}{3}\right\rfloor~$ will evaluate to $~\displaystyle \left(\frac{996 - 4x_1}{3}\right).$
For $~\displaystyle x_1 \in B, ~\left\lfloor \frac{1000 - 4x_1 - 3}{3}\right\rfloor~$ will evaluate to $~\displaystyle \left(\frac{997 - 4x_1}{3}\right).$
For $~\displaystyle x_1 \in C, ~\left\lfloor \frac{1000 - 4x_1 - 3}{3}\right\rfloor~$ will evaluate to $~\displaystyle \left(\frac{995 - 4x_1}{3}\right).$
For any value of $~\displaystyle y_1, ~ \left\lfloor \frac{1000 - 4x_1 - 1 - 3y_1}{2}\right\rfloor ~$ will equal:
$\left(\frac{1000 - 4x_1 - 1 - 3y_1}{2}\right)~$ when $y_1$ is odd, and
$\left(\frac{1000 - 4x_1 - 1 - 3y_1}{2}\right) - \left(\frac{1}{2}\right) ~$ when $y_1$ is even.
For $x_1 \in A, ~y_1$ will run from $(1)$ (an odd number) through $~\displaystyle \left(\frac{996 - 4x_1}{3}\right) ~$ (an even number). Therefore, for each value of $~\displaystyle x_1 \in A, ~\left(\frac{996 - 4x_1}{6}\right)~$ of the generated $y_1$ values will be even.
Therefore, for $x_1 \in A$, the overall summation will evaluate to
$$\sum_{x_1 \in A} \left[ ~\left(\sum_{y_1=1}^{\frac{996 - 4x_1}{3}} \frac{1000 - 4x_1 - 1 - 3y_1}{2}\right) ~-~ \left(\sum_{y_1=1}^{\frac{996 - 4x_1}{6}} \frac{1}{2}\right) ~\right]$$
$$= ~\sum_{x_1 \in A} \left[ ~\left(\sum_{y_1=1}^{\frac{996 - 4x_1}{3}} \frac{1000 - 4x_1 - 1 - 3y_1}{2}\right) ~-~ \frac{996 - 4x_1}{12} ~\right].$$
For $x_1 \in B, ~y_1$ will run from $(1)$ (an odd number) through $~\displaystyle \left(\frac{997 - 4x_1}{3}\right) ~$ (an odd number). Therefore, for each value of $~\displaystyle x_1 \in B, ~\left(\frac{994 - 4x_1}{6}\right)~$ of the generated $y_1$ values will be even.
Therefore, for $x_1 \in B$, the overall summation will evaluate to
$$\sum_{x_1 \in B} \left[ ~\left(\sum_{y_1=1}^{\frac{997 - 4x_1}{3}} \frac{1000 - 4x_1 - 1 - 3y_1}{2}\right) ~-~ \left(\sum_{y_1=1}^{\frac{994 - 4x_1}{6}} \frac{1}{2}\right) ~\right]$$
$$= ~\sum_{x_1 \in B} \left[ ~\left(\sum_{y_1=1}^{\frac{997 - 4x_1}{3}} \frac{1000 - 4x_1 - 1 - 3y_1}{2}\right) ~-~ \frac{994 - 4x_1}{12} ~\right].$$
For $x_1 \in C, ~y_1$ will run from $(1)$ (an odd number) through $~\displaystyle \left(\frac{995 - 4x_1}{3}\right) ~$ (an odd number). Therefore, for each value of $~\displaystyle x_1 \in C, ~\left(\frac{992 - 4x_1}{6}\right)~$ of the generated $y_1$ values will be even.
Therefore, for $x_1 \in C$, the overall summation will evaluate to
$$\sum_{x_1 \in C} \left[ ~\left(\sum_{y_1=1}^{\frac{995 - 4x_1}{3}} \frac{1000 - 4x_1 - 1 - 3y_1}{2}\right) ~-~ \left(\sum_{y_1=1}^{\frac{992 - 4x_1}{6}} \frac{1}{2}\right) ~\right]$$
$$= ~\sum_{x_1 \in C} \left[ ~\left(\sum_{y_1=1}^{\frac{995 - 4x_1}{3}} \frac{1000 - 4x_1 - 1 - 3y_1}{2}\right) ~-~ \frac{992 - 4x_1}{12} ~\right].$$
So, for a specific value of $x_1 \in A,$ the expression evaluates to
$\displaystyle \left[\left(\frac{996 - 4x_1}{3}\right)\left(\frac{999 - 4x_1}{2}\right) - \frac{3}{2}\left(\frac{996-4x_1}{3}\right)\left(\frac{999-4x_1}{3}\right)\left(\frac{1}{2}\right)\right]$
$\displaystyle - ~\left(\frac{996 - 4x_1}{12}\right)$
$\displaystyle = ~\left(\frac{996 - 4x_1}{3}\right) \left[\left(\frac{999 - 4x_1}{2}\right) - \left(\frac{999 - 4x_1}{4}\right)\right]$
$\displaystyle - ~\left(\frac{996 - 4x_1}{12}\right)$
$\displaystyle = ~\left(\frac{996 - 4x_1}{3}\right) \left(\frac{999 - 4x_1}{4}\right) - ~\left(\frac{996 - 4x_1}{12}\right).$
Set $\displaystyle ~\overline{x_1} = \frac{x_1 + 3}{3} \implies x_1 = 3\overline{x_1} - 3.$
As $x_1$ steps through the elements in set $A$, $\overline{x_1}$ will run from $1$ through $83.$
So, the portion of the overall computation that is represented by $x_1 \in A$ will be
$\displaystyle = ~\sum_{\overline{x_1} = 1}^{83} \left[\frac{996 - 4(3\overline{x_1} - 3)}{3}\right] \left[\frac{999 - 4(3\overline{x_1} - 3)}{4}\right] - ~\left[\frac{996 - 4(3\overline{x_1} - 3)}{12}\right]$
$\displaystyle = ~\sum_{\overline{x_1} = 1}^{83} \left(\frac{1008 - 12\overline{x_1}}{3}\right) \left(\frac{1011- 12\overline{x_1}}{4}\right) - ~\left(\frac{1008 - 12\overline{x_1}}{12}\right)$
$\displaystyle = ~\sum_{\overline{x_1} = 1}^{83} \left[\frac{1019088 - 24228\overline{x_1} + 144\overline{x_1}^2}{12} ~-~ \frac{1008 - 12\overline{x_1} }{12}\right]$
$\displaystyle = ~\sum_{\overline{x_1} = 1}^{83} \left[84840 - 2018\overline{x_1} + 12\overline{x_1}^2\right]$
$\displaystyle = 83 \times (84840) - \frac{(2018)(83)(84)}{2} + \frac{(12)(83)(84)(167)}{6}$
$\displaystyle ~=~ 2,335,620.$
For a specific value of $x_1 \in B,$ the expression evaluates to
$\displaystyle \left[\left(\frac{997 - 4x_1}{3}\right)\left(\frac{999 - 4x_1}{2}\right) - \frac{3}{2}\left(\frac{997-4x_1}{3}\right)\left(\frac{1000-4x_1}{3}\right)\left(\frac{1}{2}\right)\right]$
$\displaystyle - ~\left(\frac{994 - 4x_1}{12}\right)$
$\displaystyle = ~\left(\frac{997 - 4x_1}{3}\right) \left[\left(\frac{999 - 4x_1}{2}\right) - \left(\frac{1000 - 4x_1}{4}\right)\right]$
$\displaystyle - ~\left(\frac{994 - 4x_1}{12}\right)$
$\displaystyle = ~\left(\frac{997 - 4x_1}{3}\right) \left(\frac{998 - 4x_1}{4}\right) - ~\left(\frac{994 - 4x_1}{12}\right).$
Set $\displaystyle ~\overline{x_1} = \frac{x_1 + 2}{3} \implies x_1 = 3\overline{x_1} - 2.$
As $x_1$ steps through the elements in set $B$, $\overline{x_1}$ will run from $1$ through $83.$
So, the portion of the overall computation that is represented by $x_1 \in B$ will be
$\displaystyle \sum_{\overline{x_1} = 1}^{83} \left(\frac{[997 - 4(3\overline{x_1} - 2)]}{3}\right) \left(\frac{[998 - 4(3\overline{x_1} - 2)]}{4}\right)$
$\displaystyle- ~\left(\frac{[994 - 4(3\overline{x_1} - 2)]}{12}\right)$
$= ~\displaystyle \sum_{\overline{x_1} = 1}^{83} \left(\frac{1005 - 12\overline{x_1}}{3}\right) \left(\frac{1006 - 12\overline{x_1}}{4}\right)$
$\displaystyle - ~\left(\frac{1002 - 12\overline{x_1}}{12}\right)$
$= ~\displaystyle \sum_{\overline{x_1} = 1}^{83} 84169 - 2010\overline{x_1} + 12\overline{x_1}^2$
$= ~\displaystyle (84169)(83) - 2010\left[\frac{(83)(84)}{2}\right] + 12\left[\frac{(83)(84)(167)}{6}\right]$
$\displaystyle =~ 2,307,815.$
For a specific value of $x_1 \in C,$ the expression evaluates to
$\displaystyle \left[\left(\frac{995 - 4x_1}{3}\right)\left(\frac{999 - 4x_1}{2}\right) - \frac{3}{2}\left(\frac{995-4x_1}{3}\right)\left(\frac{998-4x_1}{3}\right)\left(\frac{1}{2}\right)\right]$
$\displaystyle - ~\left(\frac{992 - 4x_1}{12}\right)$
$\displaystyle = ~\left(\frac{995 - 4x_1}{3}\right) \left[\left(\frac{999 - 4x_1}{2}\right) - \left(\frac{998 - 4x_1}{4}\right)\right]$
$\displaystyle - ~\left(\frac{992 - 4x_1}{12}\right)$
$\displaystyle = ~\left(\frac{995 - 4x_1}{3}\right) \left(\frac{1000 - 4x_1}{4}\right) - ~\left(\frac{992 - 4x_1}{12}\right).$
Set $\displaystyle ~\overline{x_1} = \frac{x_1 + 1}{3} \implies x_1 = 3\overline{x_1} - 1.$
As $x_1$ steps through the elements in set $C$, $\overline{x_1}$ will run from $1$ through $83.$
The portion of the overall computation that is represented by $x_1 \in C$ will be
$\displaystyle \sum_{\overline{x_1} = 1}^{83} \left(\frac{[995 - 4(3\overline{x_1} - 1)]}{3}\right) \left(\frac{[1000 - 4(3\overline{x_1} - 1)]}{4}\right)$
$\displaystyle- ~\left(\frac{[992 - 2(3\overline{x_1} - 1)]}{12}\right)$
$= ~\displaystyle \sum_{\overline{x_1} = 1}^{83} \left(\frac{999 - 12\overline{x_1}}{3}\right) \left(\frac{1004 - 12\overline{x_1}}{4}\right)$
$\displaystyle - ~\left(\frac{996 - 12\overline{x_1}}{12}\right)$
$= ~\displaystyle \sum_{\overline{x_1} = 1}^{83} 83500 - 2002\overline{x_1} + 12\overline{x_1}^2$
$= ~\displaystyle (83500)(83) - 2002\left[\frac{(83)(84)}{2}\right] + 12\left[\frac{(83)(84)(167)}{6}\right]$
$\displaystyle =~ 2,280,176.$
Total for the each of the sets $A,B,$ and $C$ is
$2,335,620 + 2,307,815 + 2,280,176 = 6,923,611.$
user2661923
- 35,619
- 3
- 17
- 39