Reallife Permutation and Combination Part 1 Permutation 1 D

Real-life Permutation and Combination

Part 1 - Permutation

1. Describe and justify a real-life situation whose outcomes can be modeled by a permutation. Specifically address why your example is a permutation by discussing non-repetition of objects and if order matters. (7 points)

2. Calculate the number of outcomes for your permutation situation described above. Show your calculations. (2 points)

Part 2 - Combination

1. Describe and justify a real-life situation whose outcomes can be modeled by a combination. Specifically address why your example is a combination by discussing non-repetition of objects and if order matters. (7 points)

2. Calculate the number of outcomes for your combination situation described above. Show your calculations. (2 points)

Solution

Permutation is the total ways of arrangements taking r items at a time out of total n given items, where the order of

arrangement is always considered. So based on this definition, we can discuss the following real life situation in which permutation is used, considering it order as well as no repetion of the items at all.

REal life situation showing permutation :There are four colleges in a city. In how many ways cann a woman      

                                                                send 3 of her sons to a college. If no two of his sons are to study in the same college?

This is an example of permulation with no repetition. Here the first son can be sent to a college in 4 ways and when the first son has been sent to a college in one of these four ways, the second son can be sent to another college in 3 ways. Hence, the first two sons can be sent to different colleges in 4 x 3 ways.

The first two sons having been sent to two of the four colleges in any one of the 4x 3 ways, the third son can be sent to any of the remaining 2 colleges.

Thus, corresponding to each of 4 x 3 ways of sending the first two sons to colleges, there are 2 ways of sending the third son. Hence the requried number of ways = 4 x 3 x 2 = 24 ways.

The following are the possible choices, if a,b,c and d are four colleges :

abc, abd, acb, acd, adb, adc, bac, bad, bca, bcd, bda, bdc, cab, cad, cba, cbd, cda, cdb, dab, dac, dba, dbc, dca, dcb.

Thus there are total 4 x 3 x 2 = 24 way.