Aski patrol unit has seven members available for duty, and two of them are to be sent to rescue an injured skier. In how many ways can two of these seven members be selected?

Now suppose the order of selection is important. How many arrangements are possible in this case?

36 non repeating, 72 repeating (AB=BA)

Step-by-step explanation:

There is a theoretical approach to solving this question. We have formulae for such types of question in ‘Permutation and Combination’ chapter but let me solve it through my approach.

There are 9 members in the unit.

we have to choose only 2 people out of them randomly.

let the members be A, B, C, D, E, F, G, H an, I

A can be chosen first with one of the rest as the second person.

Like AB, AC, AD, AE … AI ………………………(1)

by this, we can choose people in 8 ways.

we will also have 8 ways as BA, CA, DA, … IA ………..(2)

the series (2) is completely reverse of series (1)

now perform the same action taking B as first-person and people after B as the second person.

by this, we will have 7 ways

and 7 ways again in reverse mode.

The same process with C, D, E and other members

we get

8*2 = 16 ways with A

7*2 = 14 ways with B

6*2 = 12 ways with C

5*2 = 10 ways with D

4*2 = 8 ways with E

3*2 = 6 ways with F

2*2 = 4 ways with G

1*2 = 2 ways with H

and I can not be selected lonely because there is no person after I.

SO total ways = 16 + 14 +12 + 10 + 8 + 6 + 4 + 2 = 72 ways

72 will be the answer if the selection of A first and B second is different from the selection of B first and A second.

If AB = BA then the answer will be 72/2 = 36

(It is my approach and easier way for me than the traditional formulae)

what graph

step-by-step explanation: