Math Problem Statement

If I divide 11 members into four groups and select atleast one member from each group and I need result in the sum of all possible combinations is select any 7 out of 11 format

Solution

To solve this problem, we need to divide 11 members into four groups, with at least one member in each group. Then, we want to determine the sum of all possible combinations that result in selecting 7 members out of the 11.

Here’s a step-by-step breakdown:

  1. Partition the 11 members into four groups G1,G2,G3,G4G_1, G_2, G_3, G_4, where each group has at least one member. This is a classic combinatorial problem of distributing objects (members) into bins (groups) with restrictions.

    • Let’s assign the group sizes as x1,x2,x3,x4x_1, x_2, x_3, x_4 such that: x1+x2+x3+x4=11withx1,x2,x3,x41x_1 + x_2 + x_3 + x_4 = 11 \quad \text{with} \quad x_1, x_2, x_3, x_4 \geq 1 Since each group must have at least one member, we subtract 1 from each variable and solve the new equation: (x1+x2+x3+x4=7)wherex1,x2,x3,x40(x_1' + x_2' + x_3' + x_4' = 7) \quad \text{where} \quad x_1', x_2', x_3', x_4' \geq 0 This is a stars-and-bars problem, which counts the number of ways to partition 7 members into 4 groups without restrictions. The solution is: (7+4141)=(103)=120(ways to partition 7 into 4 groups)\binom{7 + 4 - 1}{4 - 1} = \binom{10}{3} = 120 \quad \text{(ways to partition 7 into 4 groups)}
  2. Selecting 7 members out of 11: Now that we know the number of ways to partition 7 members into 4 groups, we can choose which 7 members to select from the 11. The total number of ways to select 7 members from 11 is: (117)=(114)=330\binom{11}{7} = \binom{11}{4} = 330

  3. Combining the results: To get the final result, we multiply the number of ways to partition 7 members into 4 groups (120) by the number of ways to select 7 members from 11 (330). The total number of combinations is: 120×330=39,600120 \times 330 = 39,600

Thus, the total number of combinations in this scenario is 39,600.

Would you like more details or clarification on any part of this?

Here are 5 related questions for further exploration:

  1. How do you solve the stars-and-bars combinatorics problem in other situations?
  2. What are the general rules for partitioning objects into groups with constraints?
  3. How can we modify this approach if the group sizes have different constraints?
  4. How do combinations work when selecting members from a set with repeated elements?
  5. How can we apply generating functions to solve combinatorial problems like this?

Tip: In combinatorics, mastering the stars-and-bars method is key for handling problems involving distributions of indistinguishable objects into distinct groups.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Stars and Bars Theorem
Combinations

Formulas

Partition formula for distributing members into groups: (x1' + x2' + x3' + x4' = 7, where x1', x2', x3', x4' ≥ 0)
Combinations formula: C(n, k) = n! / [k!(n-k)!]
Total combinations: 120 * 330 = 39,600

Theorems

Stars and Bars Theorem
Combinatorial Selection

Suitable Grade Level

Undergraduate or Advanced High School (Grades 11-12)