Question

In a high school, a committee has to be formed from a group of 6 boys \(M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6}\) and 5 girls \(G_{1}, G_{2}, G_{3}, G_{4}, G_{5}\) (i) Let \(\alpha_{1}\) be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 boys and 2 girls. (ii) Let \(\alpha_{2}\) be the total number of ways in which the committee can be formed such that the committee has at least 2 members, and having an equal number of boys and girls. (iii) Let \(\alpha_{3}\) be the total number of ways in which the committee can be formed such that the committee has 5 members, at least 2 of them being girls. (iv) Let \(\alpha_{4}\) be the total number of ways in which the committee can be formed such that the committee has 4 members, having at least 2 girls and such that both \(M_{1}\) and \(G_{1}\) are NOT in the committee together. LIST-I P. The value of \(\alpha_{1}\) is Q. The value of \(\alpha_{2}\) is \(\mathbf{R}\). The value of \(\alpha_{3}\) is \(\mathrm{S}\). The value of \(\alpha_{4}\) is LIST-II 1\. 136 2\. 189 3\. 192 4\. 200 5\. 381 6\. 461

Short answer

The values of \(\alpha_{1}\), \(\alpha_{2}\), \(\alpha_{3}\), and \(\alpha_{4}\) are 150, 381, 200, and 189 respectively, which correspond to P=4, Q=5, R=3, and S=2.

Step-by-step solution

- Calculate \(\alpha_{1}\)

To find the total number of ways a committee with exactly 3 boys and 2 girls can be formed, we use combinations. Select 3 boys from 6: \(\text{Combination of boys} = \binom{6}{3}\). Select 2 girls from 5: \(\text{Combination of girls} = \binom{5}{2}\). Multiply these combinations to get \(\alpha_{1}\): \(\alpha_{1} = \binom{6}{3} \times \binom{5}{2}\).

- Calculate \(\alpha_{2}\)

The committee can have 2, 3, 4, 5, or 6 members with an equal number of boys and girls. Calculate combinations for each equalled number case and sum them: \(\alpha_{2} = \binom{6}{1}\binom{5}{1} + \binom{6}{2}\binom{5}{2} + \binom{6}{3}\binom{5}{3}\).

- Calculate \(\alpha_{3}\)

The committee can have 5 members with at least 2 girls. First calculate the total combinations with exactly 2, 3, 4, and 5 girls and sum them: \(\alpha_{3} = \binom{6}{3}\binom{5}{2} + \binom{6}{2}\binom{5}{3} + \binom{6}{1}\binom{5}{4} + \binom{6}{0}\binom{5}{5}\).

- Calculate \(\alpha_{4}\)

The committee has 4 members with at least 2 girls and without both \(M_{1}\) and \(G_{1}\) together. Calculate separately for cases: both \(M_{1}\) and \(G_{1}\) absent, only \(M_{1}\) absent, and only \(G_{1}\) absent. Add the results: \(\alpha_{4} = \binom{5}{2}\binom{4}{2} + \binom{5}{3}\binom{4}{1} + \binom{4}{4}\binom{5}{0}\).

Key concepts

Permutations and Combinations

Permutations and combinations are fundamental concepts in combinatorics, the branch of mathematics that deals with counting. In the context of forming committees or selecting a group from a larger set, understanding these concepts is essential.

Permutations refer to arrangements of items where the order matters. For instance, the way books are arranged on a shelf is a permutation of the books. In contrast, combinations are selections where the order does not matter, such as choosing a group of students for a project from a class. This is what we use when we calculate the number of possible committees.

Mathematically, combinations are represented using binomial coefficients, often denoted as \(\binom{n}{r}\), which is read as 'n choose r'. This gives the number of ways to choose r elements from a set of n distinct elements. To reinforce this concept, let's take a closer look at the provided exercises and solutions.

\textbf{Example:}\
To determine \(\alpha_1\), the number of ways to form a committee with exactly 3 boys out of 6 and 2 girls out of 5, we use the formula for combinations. For boys, we calculate \(\binom{6}{3}\), and for girls, \(\binom{5}{2}\), and then multiply them together because the selection of boys and girls are independent events. This multiplication rule is a core principle in combinatorics that applies when combining different independent choices.

Committee Formation

Committee formation is a common application of combinatory concepts, and it demonstrates the practical use of permutations and combinations. When forming a committee, the selection process usually does not consider the order in which members are chosen, which turns the scenario into a problem of finding combinations.

Understanding how to tackle various selection criteria is key to solving these problems. For instance, in our exercise, different conditions change the way we count the number of possible committees:

\textbf{Different Selections:}

\textbf{At least 'x' members:} Problems like \(\alpha_2\) and \(\alpha_3\) involve a condition about having 'at least' a certain number of members. These problems typically require calculating multiple cases separately and then summing them to get the total count.

\textbf{Excluding specific members:} In \(\alpha_4\), there's an additional rule prohibiting certain individuals from being chosen together, adding a layer of complexity. Here, cases need to be considered separately: first where both excluded members are absent, then where each one is absent, calculating combinations accordingly for the remaining choices.

As seen in the solutions, the formation of committees can have many variations and understanding how to apply the rules of combination correctly to each unique situation is paramount for students to master these problems.

Binomial Theorem

The Binomial Theorem is a powerful tool in algebra that relates to combinatorics. It provides a systematic method for expanding expressions that are raised to a power, such as \( (x + y)^n \). The coefficients of the expanded terms are precisely the binomial coefficients that appear in combination problems.

The connection between the binomial theorem and combinatorics is particularly evident in problems involving 'choose' numbers or combinations. Each term in the expansion of a binomial expression corresponds to a binomial coefficient, which resembles the 'choose' function in a combinatorial setting.

For example, the coefficient of the \( x^3y^2 \) term when expanding \( (x+y)^5 \) is the same as the number of ways to choose 3 objects from 5, which is \( \binom{5}{3} \). This correlation is not just a mathematical curiosity; it provides a direct link between algebraic expansion and counting principles, highlighting the consistent patterns found within mathematics.

In summary, the binomial theorem is not only key in algebra but also plays a crucial role in solving combinatorial problems. Familiarity with this theorem can greatly facilitate the understanding of concepts like committee formation, where it can be used to predict the number of outcomes based on certain conditions.