Question

The no of ways in which 10 persons can go in two cars so that there may be 5 in each car, supposing that two particular persons will not go in the same car is: (a) \((1 / 2)\left({ }^{10} \mathrm{C}_{5}\right)\) (b) \((1 / 2)\left({ }^{8} \mathrm{C}_{5}\right)\) (c) \(2\left({ }^{8} \mathrm{C}_{4}\right)\) (d) \({ }^{8} \mathrm{C}_{4}\)

Short answer

The short answer to the given problem is: \[\binom{10}{5} - 2\binom{8}{4}\] And the correct option is (c).

Step-by-step solution

Identify the possible cases without the constraint

First, we need to find the total number of ways to divide the 10 persons into two groups of 5 each, without the constraint that the two specific persons cannot go in the same car. This can be done using combinations. We need to choose 5 persons out of 10 to go in the first car. This can be done in \(\binom{10}{5}\) ways.

Subtract the cases which include the constraint

Now, we need to count the number of cases when the two specific persons are in the same car. We can think of these two persons as a single entity and the problem becomes choosing 4 other persons out of the remaining 8 to go in the same car with the pair. This can be done in \(\binom{8}{4}\) ways. Since there are two cars, we need to multiply this value by 2.

Calculate the final number of ways

To find the final number of ways, we need to subtract the cases calculated in Step 2 from the total number of ways found in Step 1. So the answer to the problem is: \[\binom{10}{5} - 2\binom{8}{4}\]

Check which option matches our answer

Looking at the given options, we can see that option (c) is the same as our answer: \[\textbf{(c)}\; 2\left({ }^{8} \mathrm{C}_{4}\right)\] So, the correct answer is option (c).

Key concepts

Permutations and Combinations

Permutations and combinations are fundamental concepts in combinatorics that help us count how objects can be arranged or selected. In permutations, the order of the items matters, while in combinations, the order does not matter.
For example, when picking players for a sports team, if the order of selection doesn't matter, we use combinations. But if you're arranging books on a shelf, where the order is important, permutations are used.

• Combinations: Used when the order doesn't matter, calculated with the formula \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\).
• Permutations: Used when the order does matter, calculated with the formula \(P(n, r) = \frac{n!}{(n-r)!}\).

In the original problem, we used combinations because we were selecting people to fill spots across two cars where the order in which they were chosen didn’t affect the outcome.

Constraints in Combinatorics

Constraints add specific conditions to a combinatorial problem, making it more complex to solve. In the given problem, the constraint is that two specific individuals cannot share the same car. Such constraints mean we cannot directly apply typical permutation or combination formulas without modification.
To work around constraints:

• Identify all possible formations without considering the constraint to get a baseline understanding.
• Next, calculate cases that violate the constraint separately, often by treating them as a different combination problem.
• Subtract the violating cases from the total, like we did when subtracting the cases where the two specific individuals sit together.

Constraints require careful subtraction or adjustment strategies, ensuring no overcounting or undercounting occurs.

Binomial Coefficient

The binomial coefficient \(\binom{n}{k}\) is a cornerstone of combinatorics, representing how many ways you can choose \(k\) elements from set of \(n\) elements without considering the order. It's also a key player in the binomial theorem, which expands powers of binomials.

In the exercise, the binomial coefficient \(\binom{10}{5}\) is used to determine the initial number of ways to split 10 people. This provides insights into how many ways you can choose 5 people out of 10. The second part, \(\binom{8}{4}\), helps calculate the violating constraint cases by choosing additional people to join the pair.

• The formula for calculating a binomial coefficient is \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\).
• It's vital for evaluating terms where the order of selection doesn't alter the result, making it perfect for our problem scenario.

Understanding how to apply these coefficients helps unlock solutions to complex combinatorial problems.

Combinatorial Reasoning

Combinatorial reasoning involves forming logical arguments and solutions that involve calculating combinations, permutations, and arrangement of sets. It's not just about plugging numbers into formulas but understanding how different elements interact and relate within constraints.
To enhance combinatorial reasoning:

• Formulate the problem by breaking it down logically, often by creating small examples to visualize possible arrangements.
• Use a subtraction approach to remove overcounted cases that violate constraints or simplifying assumptions.
• Practice iterative problem solving by apply step-by-step logical reasoning as demonstrated in the step-by-step solution provided, perfecting each step before moving to the next.

The ability to think combinatorially is critical in solving problems accurately, ensuring that the logic is sound and all potential scenarios have been considered.