Short Tricks to Solve Dice Sum Problem (2)

Duration: 12 min

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This educational video is a mathematics lecture that explains how to solve a classic combinatorics problem: finding the number of ways to get a sum of 10 when three dice are thrown. The video begins with a title slide that introduces the topic of permutations and combinations, setting the context for the problem. The instructor then presents the problem statement: 'If three dice are thrown, find the number of ways to get sum 10?'. The core of the lesson involves a methodical approach using the stars and bars technique, but with constraints. The variables x1, x2, and x3 are defined as the values on the three dice, with the equation x1 + x2 + x3 = 10. The constraint 1 ≤ xi ≤ 6 is applied to each variable, as a die cannot show a value less than 1 or more than 6. The instructor demonstrates the standard method for solving such problems by first transforming the variables to remove the lower bound (by setting xi' = xi - 1), which leads to the equation x1' + x2' + x3' = 7. The total number of non-negative integer solutions is calculated as C(7+3-1, 3-1) = C(9,2). To account for the upper bound constraint, the instructor uses the principle of inclusion and exclusion. The method involves subtracting the cases where one or more variables exceed 6. For example, if x1 > 6, a substitution x1'' = x1 - 6 is made, leading to a new equation x1'' + x2' + x3' = 1. The number of solutions for this case is C(1+3-1, 3-1) = C(3,2). Since there are three dice, this case is multiplied by 3. The final answer is the total number of solutions minus the invalid cases: C(9,2) - 3*C(3,2) = 36 - 9 = 27. The video concludes with a general formula for the number of ways to get a sum n with three dice, which is C(n-1,2) - 3*C(n-7,2) for n from 7 to 12, and a final 'Thanks for watching' screen.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with a title slide that introduces the topic of permutations and combinations. The slide is divided into four quadrants, each with a visual metaphor: a green craps table with dice for 'PERMUTATION', a basket of fruit for 'COMBINATION', a pizza for 'COMBINATION', and two men in a business meeting for 'PERMUTATION'. This is followed by a presentation slide titled 'Confused' which poses the question, 'Should I unlock with Permutation or Combination?'. The slide features a cartoon detective with a magnifying glass, standing between two locked doors labeled 'Permutation' and 'Combination', visually representing the confusion between the two concepts. The instructor, Yash Jain, is visible in a small window in the bottom right corner. The slide also includes the 'Knowledge Gate Educator' logo and branding.

  2. 2:00 – 5:00 02:00-05:00

    The video transitions to a new slide titled 'Dice Sum Problem'. The instructor presents the core problem: 'If three dice are thrown, find the number of ways to get sum 10?'. The instructor begins to set up the mathematical model by defining the variables x1, x2, and x3 as the values on the three dice. The equation x1 + x2 + x3 = 10 is written on the screen. The constraints 1 ≤ x1 ≤ 6, 1 ≤ x2 ≤ 6, and 1 ≤ x3 ≤ 6 are also written, reflecting the possible values on a standard die. The instructor explains that this is a problem of finding the number of integer solutions to this equation under the given constraints.

  3. 5:00 – 10:00 05:00-10:00

    The instructor explains the method to solve the problem. He first removes the lower bound constraint by substituting xi' = xi - 1, which transforms the equation to x1' + x2' + x3' = 7, where xi' ≥ 0. He then uses the stars and bars method to find the total number of non-negative integer solutions, which is C(7+3-1, 3-1) = C(9,2). He then addresses the upper bound constraint. He explains that the total count includes invalid cases where one of the variables is greater than 6. He demonstrates the inclusion-exclusion principle by calculating the number of invalid cases. For example, if x1 > 6, he substitutes x1'' = x1 - 6, leading to the equation x1'' + x2' + x3' = 1. The number of solutions for this is C(1+3-1, 3-1) = C(3,2). Since there are three dice, this case is multiplied by 3. The final answer is the total number of solutions minus the invalid cases: C(9,2) - 3*C(3,2) = 36 - 9 = 27.

  4. 10:00 – 11:47 10:00-11:47

    The instructor generalizes the solution. He writes the formula for the number of ways to get a sum n with three dice as C(n-1,2) - 3*C(n-7,2) for n from 7 to 12. He explains that this formula accounts for the total number of solutions and subtracts the cases where one of the dice shows a value greater than 6. The video concludes with a black screen displaying the text 'THANKS FOR WATCHING' in an orange and white box.

The video provides a clear, step-by-step tutorial on solving a combinatorics problem involving dice. It begins by establishing the context of permutations and combinations, then focuses on a specific problem. The core of the lesson is the application of the stars and bars method to find the number of integer solutions to an equation, with a crucial emphasis on handling constraints. The instructor demonstrates the principle of inclusion and exclusion to subtract invalid cases where a die value exceeds 6. The progression from setting up the problem to applying the formula and generalizing the result provides a comprehensive understanding of the method, making it a valuable resource for students learning combinatorics.