Probabaility of having certain number of parents in a committee

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SUMMARY

The committee formation problem involves selecting 5 members from a group of 6 parents, 2 teachers, and a principal, with the condition that no more than 4 parents can be included. The correct combinations are calculated using binomial coefficients, specifically focusing on scenarios with 2, 3, or 4 parents. The total number of valid combinations is determined to be 120, as confirmed by the calculations of (6C2 x 7C3) + (6C3 x 6C2) + (6C4 x 5C1).

PREREQUISITES
  • Understanding of combinatorial mathematics, specifically binomial coefficients.
  • Familiarity with the concept of committee formation and selection problems.
  • Knowledge of basic probability principles.
  • Ability to perform calculations involving combinations, denoted as nCr.
NEXT STEPS
  • Study the principles of combinatorial mathematics, focusing on binomial coefficients.
  • Learn about advanced counting techniques in probability theory.
  • Explore real-world applications of committee selection problems in organizational settings.
  • Practice similar problems involving combinations and permutations to reinforce understanding.
USEFUL FOR

Students studying combinatorial mathematics, educators teaching probability concepts, and anyone interested in solving selection problems in statistics.

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Homework Statement



A committee of 5 members is formed from 6 parents , 2 teachers, and a principal, In how many ways, can this committee be formed if the committee consists of not more than 4 parents?

my working is 1 parent + 2 parents + 3 parents +4 parents

which is (6C1 x 8C4) + (6C2 x 7C3) +(6C3x 6C2)+ (6C4 x 5C1)

but the ans given is only 120

Homework Equations





The Attempt at a Solution

 
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desmond iking said:

Homework Statement



A committee of 5 members is formed from 6 parents , 2 teachers, and a principal, In how many ways, can this committee be formed if the committee consists of not more than 4 parents?

my working is 1 parent + 2 parents + 3 parents +4 parents

which is (6C1 x 8C4) + (6C2 x 7C3) +(6C3x 6C2)+ (6C4 x 5C1)

but the ans given is only 120

Homework Equations





The Attempt at a Solution


Since there are only 3 non-parents your choices for number of parents are 2,3, and 4. Once you have counted the parents, the others must be chosen from the non-parents. Try again.
 
LCKurtz said:
Since there are only 3 non-parents your choices for number of parents are 2,3, and 4. Once you have counted the parents, the others must be chosen from the non-parents. Try again.

Does the identities of the parents and teachers matter? In other words, would a 3-parent committee having parents Smith, Jones and Chakravarty be counted separately from a 3-parent committee having parents Smith, Jones and Salam? Ditto for the teachers.
 

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