Problem 90 - Algebra and Trigonometry Real Mathematics, Real People

Question

Without calculating, determine whether the value of \(_{n} P_{r}\) is greater than the value of \(_{n} C_{r}\) for the values of \(n\) and \(r\) given in the table. Complete the table using yes (Y) or no (N). Is the value of \(_{n} P_{r}\) always greater than the value of \(_{n} C_{r} ?\) Explain.

Step-by-Step Solution

Answer

Yes, the value of \( _{n} P_{r} \) is always greater than the value of \( _{n} C_{r} \), because permutations (which consider order of arrangement) will always provide more possibilities than combinations (where order does not matter).

Step 1: Definition of \( _{n} P_{r} \)and \( _{n} C_{r} \)

Permutation (represented as \( _{n} P_{r} \)) is the number of different ways that \(r\) items can be chosen from \(n\) items where the order of selection matters. Combination (represented as \( _{n} C_{r} \)) is the number of ways that \(r\) items can be chosen from \(n\) items where the order does not matter.

Step 2: Comparison of \( _{n} P_{r} \)and \( _{n} C_{r} \)

Since permutations take into account the order of the items, they always yield more possibilities than combinations, where the order does not matter. Hence, \( _{n} P_{r} \) is always greater than \( _{n} C_{r} \) for the same values of \(n\) and \(r\).

Step 3: Explanation of The Result

Take an example to demonstrate this. For any given numbers of \(n\) and \(r\), say 5 and 3, the number of permutations (83 ways) is indeed larger than the number of combinations (10 ways). This is because the permutations count different sequences separately while combinations count all sequences as one if they have the same elements. Therefore, regardless of the given values of \(n\) and \(r\), \( _{n} P_{r} \) is always greater than \( _{n} C_{r} \).