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Tips & Tricks: Permutation & Combination - 1 Video Lecture | Quantitative Aptitude for SSC CGL
FAQs on Tips & Tricks: Permutation & Combination - 1 Video Lecture - Quantitative Aptitude for SSC CGL
| 1. What is the fundamental difference between permutation and combination? |  |
Ans. The fundamental difference lies in the order of selection. Permutation refers to the arrangement of items where the order matters, while combination refers to the selection of items where the order does not matter. For example, arranging the letters A, B, and C in different ways (ABC, ACB, BAC, etc.) is a permutation, whereas selecting 2 letters from A, B, and C (AB, AC, BC) is a combination.
| 2. How can I calculate the number of permutations of a set of n distinct objects? | |
Ans. The number of permutations of a set of n distinct objects can be calculated using the factorial function, denoted as n!. The formula is n! = n × (n-1) × (n-2) × ... × 1. For instance, for 4 distinct objects (A, B, C, D), the number of permutations is 4! = 4 × 3 × 2 × 1 = 24.
| 3. What is the formula for calculating combinations, and how is it different from permutations? | |
Ans. The formula for calculating combinations is given by nCr = n! / [r! × (n - r)!], where n is the total number of items, and r is the number of items to choose. This formula considers only the selection without regard to the order, which differentiates it from the permutation formula. For example, in choosing 2 items from 4 (A, B, C, D), the combinations are AB, AC, AD, BC, BD, and CD, totaling 6 combinations.
| 4. In what scenarios would one use permutations instead of combinations? | |
Ans. One would use permutations in scenarios where the arrangement of items is significant. For instance, if the task involves forming different sequences of a set of items, such as seating arrangements, race placements, or password generation, permutations are applicable. Conversely, if the focus is solely on the selection of items without regard to arrangement, combinations would be appropriate.
| 5. Can you provide an example where both permutations and combinations are used in a single problem? | |
Ans. Certainly! Consider a scenario where a committee of 3 members is to be formed from a group of 5 people (A, B, C, D, E). First, we would use combinations to select the members, which can be calculated as 5C3 = 10 ways. Then, if we need to arrange these selected members (say in a specific order), we would use permutations, calculating the arrangements of the chosen 3 members, which would be 3! = 6. Thus, the total number of ways to select and arrange would be 10 (combinations) × 6 (permutations) = 60 different outcomes.
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