What are the preferred steps to solve an exercise that involves finding all possible combinations/solutions for a string with constraints?

Question

I have an exercise divided in two parts:

a.) How many ( $x \in \mathbb{Z}$ ), with ($ 104050607080 \leq x \leq 908070605040$ ), can be formed using the digits of ( $106506506503$ ), such that ( $x$ ) is divisible by $20$ and contains the string "$036$" as a substring?

b.) How many solutions are there to the equation ( $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 = 8004$ ), where ( $x_1, \ldots, x_8 \in \mathbb{Z}$ ) and ( $x_1, \ldots, x_8 \geq 0$ ), with the following constraints:

• ( $x_1 \geq 49$ ),
• ( $300 \leq x_3 \leq 399$ ),
• ( $550 \leq x_5 \leq 749$ ),
• ( $x_6 \geq 4$ ),
• ( $x_1 + x_7 + x_8 = 350$ ),
• ( $2x_1 + x_3 + x_5 \geq 950$ )?

Since I have no idea on where to start (except doing some case work for the specified constraints) I would ask if there is someone kind me enough to show me how to solve this kind of exercise. For part b. I think it's more straightforward. I would need to find all the possible values for each $x_1...x_8$ that satisfy the constraints and then multiply the number of possible choices for each $x$. Is that correct?

Edit:

• Please note that substring can only appear once.
• Please note that the leftmost digit of any positive integer is not permitted to equal 0.
• Please also note that I am not allowed to use computational tools otherwise I would have solved this exercise in C++. For instance, for part b I have been suggested to use inclusion-exclusion along with stars and bars.

I am doing, very slowly though, some progress on part b of the exercise. This is what I've done so far by following the example of another exercise solution:

Step 1: Transform the Variables

Transform the variables to eliminate the minimum bounds:

• Let $~ y_1 = x_1 - 49 $
• Let $~ y_3 = x_3 - 300 $
• Let $~ y_5 = x_5 - 550 $
• Let $~ y_6 = x_6 - 4 $

This transforms the equation to: $$ (y_1 + 49) + x_2 + (y_3 + 300) + x_4 + (y_5 + 550) + (y_6 + 4) + x_7 + x_8 = 8004 $$

Simplifying, we get: $$ y_1 + x_2 + y_3 + x_4 + y_5 + y_6 + x_7 + x_8 = 7101 $$

Step 2: Count Solutions without Constraints

The number of (non-negative integer) solutions to $~ y_1 + x_2 + y_3 + x_4 + y_5 + y_6 + x_7 + x_8 = 7101 $ is: $$ \binom{7101 + 7}{7} = \binom{7108}{7} $$

Step 3: Apply Constraints Using Inclusion-Exclusion

Constraint: $ ~x_1 + x_7 + x_8 = 350 $

The transformed constraint is: $$ y_1 + 49 + x_7 + x_8 = 350 $$ $$ y_1 + x_7 + x_8 = 301 $$ The number of solutions is: $$ \binom{301 + 2}{2} = \binom{303}{2} $$

Constraint: $~ 300 \leq x_3 \leq 399 $

The transformed constraint is: $$ 0 \leq y_3 \leq 99 $$

Constraint: $~ 550 \leq x_5 \leq 749 $

The transformed constraint is: $$ 0 \leq y_5 \leq 199 $$

Constraint: $~ x_6 \geq 4 $

The transformed constraint is: $$ y_6 \geq 0 $$

Constraint: $~ 2x_1 + x_3 + x_5 \geq 950 $

The transformed constraint is: $$ 2(y_1 + 49) + (y_3 + 300) + (y_5 + 550) \geq 950 $$ $$ 2y_1 + y_3 + y_5 \geq 2 $$

Inclusion-Exclusion Principle

We need to count the number of solutions to:
$$ y_1 + x_2 + y_3 + x_4 + y_5 + y_6 + x_7 + x_8 = 7101 $$ that satisfy the constraints using inclusion-exclusion.

1. Initial Count: $$ \binom{7108}{7} $$

2. Subtract invalid solutions where constraints are not met:

Constraint 1: $$ y_1 + x_7 + x_8 = 301 $$

The number of solutions is:
$$ \binom{303}{2} $$

Constraint 2: $~ 0 \leq y_3 \leq 99 $

For $~ y_3 $: $$ \sum_{k=0}^{99} \binom{7101 - k + 6}{6} $$

Constraint 3: $~ 0 \leq y_5 \leq 199 $

For $~ y_5 $: $$ \sum_{k=0}^{199} \binom{7101 - k + 6}{6} $$

Combined Constraint: $~ 2y_1 + y_3 + y_5 \geq 2 $

We use inclusion-exclusion to count solutions satisfying $~ 2y_1 + y_3 + y_5 \geq 2 $:

$$ \sum_{k=0}^{99} \sum_{j=0}^{199} \binom{7101 - k - j + 5}{5} $$

So, the inclusion-exclusion principle gives us: $$ N = \binom{7108}{7} - \sum_{k=0}^{99} \binom{7101 - k + 6}{6} - \sum_{k=0}^{199} \binom{7101 - k + 6}{6} + \sum_{k=0}^{99} \sum_{j=0}^{199} \binom{7101 - k - j + 5}{5} $$

Answer 1

This answer will use both Inclusion-Exclusion and Stars and Bars. See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

For Stars and Bars theory, see this article and this article.

$\underline{\text{Part a}}$

How many $~x \in \mathbb{Z}~$, with $~104050607080 \leq x \leq 908070605040,~$ can be formed using the digits of $~106506506503,~$ such that $~x~$ is divisible by $~20~$ and contains the string "$036$" as a substring?

Let the multiset $~A~$ denote $~\{0,0,0,0,1,3,5,5,5,6,6,6\}.~$ Since the lower and upper bounds for $~x~$ are each $~12~$ digit numbers, and $~A~$ contains exactly $~12~$ elements, when constructing $~x,~$ each element in $~A~$ must be used exactly once.

With respect to the constraints:

• The upper bound can be ignored, because $~A~$ does not contain a $~9.$
• The lower bound will be violated if and only if $~x~$ begins either "0", or "100" or "103".
• Since $~A~$ contains only one $~3,~$ the "036" string can only occur once. Reading the digit positions from left to right, the "036" string must start in one of the digit positions $~1~$ through $~10~$ inclusive.
• The divisibility by $~20,~$ constraint will be violated if and only if $~x~$ does not end in either "00" or "60".

Using Inclusion-Exclusion, let the set $~S~$ denote the set of all $~12~$ digit numbers constructed from the set $~A,~$ that contain the string "036". Let $~S_1~$ denote the subset of $~S~$ where the lower bound is violated. Let $~S_2~$ denote the subset of $~S~$ where the divisibility by $~20~$ is violated.

Then, the desired computation is

$$|~S~| - |S_1 \cup S_2| \\ = | ~S~| - |~S_1~| - |~S_2~| + |~ S_1 \cap S_2~|.\tag1 $$

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$\underline{\text{Computation of} ~| ~S ~|}$

The "036" can occur in any of $~10~$ positions. For each such position, the remaining elements of $~\{0,0,0,1,5,5,5,6,6\}~$ need to be distributed among the $~9~$ remaining positions. Therefore,

$$| ~S~| = 10 \times \frac{9!}{3! ~3! ~2!} = 50400.$$

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$\underline{\text{Computation of} ~| ~S_1 ~|}$

There are $~5~$ separate cases to consider.

• The number begins "036".
  The computation is $~\displaystyle \frac{9!}{3! ~3! ~2!} = 5040.$

• The number begins "0", with the "036" string occurring in one of positions 2 through 10 inclusive.
  The computation is $~\displaystyle 9 \times \frac{8!}{3! ~2! ~2!} = 15120.$

• The number begins "1036".
  The computation is $~\displaystyle \frac{8!}{3! ~3! ~2!} = 560.$

• The number begins "10036".
  The computation is $~\displaystyle \frac{7!}{3! ~2! ~2!} = 210.$

• The number begins "100", with the "036" string occurring in one of positions 4 through 10 inclusive.
  The computation is $~\displaystyle 7 \times \frac{6!}{3! ~2!} = 420.$

Therefore,

$$|~S_1~| = 5040 + 15120 + 560 + 210 + 420 = 21350.$$

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$\underline{\text{Computation of} ~| ~S_2 ~|}$

There are $~3~$ separate cases to consider.

• The number ends "036".
  The computation is $~\displaystyle \frac{9!}{3! ~3! ~2!} = 5040.$

• The number has "036" in position 9, but does not end in "0".
  The computation is $~\displaystyle \frac{9!}{3! ~3! ~2!} - \frac{8!}{3! ~2! ~2!} = 3360.$

• The number has "036" in one of positions 1 thru 8 inclusive, but does not end in either "60" or "00".
  The computation is
  $~\displaystyle 8 \times \left[ ~\frac{9!}{3! ~3! ~2!} - \frac{7!}{3! ~2!} - \frac{7!}{3! ~2!} ~\right] = 33600.$

Therefore,

$$|~S_2~| = 5040 + 3360 + 33600 = 42000.$$

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$\underline{\text{Computation of} ~| ~S_1 \cap S_2 ~|}$

There are $~9~$ cases to consider.

• The number begins "036" but does not end in either "00" or "60".
  The computation is $~\displaystyle \frac{9!}{3! ~3! ~2!} - \frac{7!}{3! ~2!} - \frac{7!}{3! ~2!} = 4200.$

• The number begins "0" and ends in "036".
  The computation is $~\displaystyle \frac{8!}{3! ~2! ~2!} = 1680.$

• The number begins "0", has "036" starting in position 9, but does not end in "0".
  The computation is $~\displaystyle \frac{8!}{3! ~2! ~2!} - \frac{7!}{3! ~2!} = 1260.$

• The number begins "0", has "036" starting in one of positions 2 through 8, but does not end in either "00" or "60".
  The computation is
  $\displaystyle 7 \times \left[ ~\frac{8!}{3! ~2! ~2!} - \frac{6!}{3! ~2!} - \frac{6!}{3!}~\right] = 10500.$

• The number begins "1036" but does not end in either "00" or "60".
  The computation is $~\displaystyle \frac{8!}{3! ~3! ~2!} - \frac{6!}{3! ~2!} - \frac{6!}{3! ~2!} = 440.$

• The number begins "10036" but does not end in either "00" or "60".
  The computation is $~\displaystyle \frac{7!}{3! ~2! ~2!} - \frac{5!}{3! ~2!} - \frac{5!}{3!} = 180.$

• The number begins "100" and ends in "036".
  The computation is $~\displaystyle \frac{6!}{3! ~2!} = 60.$

• The number begins "100", has "036" starting in position 9, but does not end in "0".
  The computation is $~\displaystyle \frac{6!}{3! ~2!} - \frac{5!}{3! ~2!} = 50.$

• The number begins "100", has "036" starting in one of positions 4 through 8, but does not end in "60".
  The computation is
  $\displaystyle 5 \times \left[ ~\frac{6!}{3! ~2!} - \frac{4!}{3!}~\right] = 280.$

Therefore,

$$|~S_1 \cap S_2 ~| \\= 4200 + 1680 + 1260 + 10500 + 440 + 180 + 60 + 50 + 280 \\= 18650.$$

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$\underline{\text{Final Computation}}$

$$| ~S~| - |~S_1~| - |~S_2~| + |~ S_1 \cap S_2 ~| \\= 50400 - [21350 + 42000] + 18650 = 5700.$$

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$\underline{\text{Part b}}$

b.) How many solutions are there to the equation ( $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 = 8004$ ), where ( $x_1, \ldots, x_8 \in \mathbb{Z}$ ) and ( $x_1, \ldots, x_8 \geq 0$ ), with the following constraints:

• ( $x_1 \geq 49$ ),
• ( $300 \leq x_3 \leq 399$ ),
• ( $550 \leq x_5 \leq 749$ ),
• ( $x_6 \geq 4$ ),
• ( $x_1 + x_7 + x_8 = 350$ ),
• ( $2x_1 + x_3 + x_5 \geq 950$ )?

The first portion of your analysis for part (b) looks good. However, I would organize things differently.

Set

• $y_1 = x_1 - 49 \implies 0 \leq y_1.$
• $y_3 = x_3 - 300 \implies 0 \leq y_3 \leq 99.$
• $y_5 = x_5 - 550 \implies 0 \leq y_5 \leq 199.$
• $y_6 = x_6 - 4 \implies 0 \leq y_6.$
• For $~i \in \{2,4,7,8\}, ~$ set $~y_i = x_i.~$

So, at this point, you are working exclusively in the variables $~y_1, \cdots, y_8,~$ where all of these variables are integers with a lower bound of $~0,~$ and only the variables $~y_3~$ and $~y_5~$ have an upper bound.

Then, the constraints to be satisfied are

• Constraint 1: $~y_1 + \cdots + y_8 = 8004 - 903 = 7101.$
• Constraint 2: $~y_1 + y_7 + y_8 = 350 - 49 = 301.$
• Constraint 3: $~2y_1 + y_3 + y_5 \geq 950 - [98 + 300 + 550] = 2.$

At this point, it is time to come up with a strategy.

Since the minimum value of $~2y_1 + y_3 + y_5~$ is $~0,~$ the number of ordered triples $~(y_1, y_3, y_5)~$ that are in violation can be individually scrutinized.

Further, Based on Constraint 2, Constraint 1 can be re-expressed as

• Constraint 4: $~y_2 + \cdots + y_6 = 7101 - 301 = 6800.$

Further, Constraint 4 and Constraint 2 share no variables. Therefore, if you temporarily ignore Constraint 3, the number of satisfying solutions to Constraint 4 and Constraint 2 are

$$\text{[Number of solutions to Constraint 4}] \\\times [\text{Number of solutions to Constraint 2}]. $$

So, my overall approach to part b will be to temporarily ignore Constraint 3, compute the corresponding product, and then deduct the number of solutions where Constraint 3 is violated. When needed, I will follow the model in this answer.

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$\underline{\text{Number of solutions to Constraint 2}}$

Constraint 2 is

• $y_1 + y_7 + y_8 =301.$
• $y_1, y_7, y_8 \in \Bbb{Z_{\geq 0}}.$

The number of solutions is

$$\binom{303}{2}.$$

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$\underline{\text{Number of solutions to Constraint 4}}$

Constraint 4, with the upper bounds included is

• $y_2 + \cdots + y_6 = 6800.$
• $y_2, \cdots, y_6 \in \Bbb{Z_{\geq 0}}.$
• $y_3 \leq 99.$
• $y_5 \leq 199.$

Let :

• $~S~$ denote the set of all solutions to the above problem, where the upper bound constraints on $~y_3~$ and $~y_5~$ are ignored.
• $~S_1~$ denote the subset of $~S~$ where $~y_3 \geq 100.$
• $~S_2~$ denote the subset of $~S~$ where $~y_5 \geq 200.$

Then, the desired computation is

$$| ~S ~| - | ~S_1~| - | ~S_2 ~| + | ~S_1 \cap S_2 ~|.$$

Following the linked model, the computation is

$$\binom{6804}{4} - \binom{6704}{4} - \binom{6604}{4} + \binom{6504}{4}.$$

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$\underline{\text{Number of combined solutions to Constraints 2 and 4}}$

The computation is

$$\binom{303}{2}$$

$$\times \left[ ~\binom{6804}{4} - \binom{6704}{4} - \binom{6604}{4} + \binom{6504}{4} ~\right].$$

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$\underline{\text{Individual} ~y_1, y_3, y_5 ~\text{Violations}}$

The constraint to be violated is

$$2y_1 + y_3 + y_5 \geq 2.$$

Since $~y_1, y_3, y_5~$ must each be non-negative integers, the following is the complete set of ordered triplets $~(y_1, y_3, y_5)~$ that are in violation:

$$\{ ~(0, 0, 0), ~(0,0,1), ~(0,1,0) ~\}.$$

So, you simply repeat all of the analysis so far, three times. once for each of the possible $~(y_1, y_3, y_5)~$ violations. You obtain three products, sum them, and then deduct that from the product already computed.

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$\underline{\text{Number of Violation Solutions to Constraint 2}}$

Here, $~y_1~$ is set to $~0.~$ So, the number of violation solutions to Constraint 2 is

$$\binom{302}{1}.$$

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$\underline{\text{Number of Violation Solutions to Constraint 4}}$

With $~y_3, y_5~$ each fixed at either $~0~$ or $~1,~$ you will have (in effect) three variables, instead of $~5.~$ Further, the sums of the three violation constraints will either be $~6800, 6799,~$ or $~6799.$

Therefore, the number of violation solutions to Constraint 4 is

$$\binom{6802}{2} + \left[ ~2 \times \binom{6801}{2} ~\right].$$

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$~\underline{\text{Total Number of Violation Solutions}}$

$$\binom{302}{1} \times \left\{ ~\binom{6802}{2} + \left[ ~2 \times \binom{6801}{2} ~\right] ~\right\}.$$

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$~\underline{\text{Final Computation.}}$

Set

$$B = \binom{303}{2}$$

$$\times \left[ ~\binom{6804}{4} - \binom{6704}{4} - \binom{6604}{4} + \binom{6504}{4} ~\right].$$

Set

$$C = \binom{302}{1}\times \left\{ ~\binom{6802}{2} + \left[ ~2 \times \binom{6801}{2} ~\right] ~\right\}.$$

Then, the final computation is

$$B - C.$$