# Meaning and Derivation of ^{n}P_{r} and ^{n}C_{r}

*Number of permutations of n different things taking r at a time =*

^{n}P_{r}In this statement, we take the following two assumptions:

- All the
*n*things are distinct (or no two things are of the same type) - Each thing is used at most once (i.e., nothing is repeated in any arrangement)

Let us assume that there are

The first box can be filled in
Hence,

*r*boxes and each of them can hold one thing. When all the*r*boxes are filled, what we have is an arrangement of*r*things taken from the given*n*things. So, each time we fill up the*r*boxes with things taken from the given*n*things, we have an arrangement of*r*things taken from the given*n*things without repetition. Hence, the number of ways in which we can fill up the*r*boxes by taking things from the given*n*things is equal to the number of permutations of*n*things taking*r*at a time.The first box can be filled in

*n*ways (because this box can be filled by any one of the*n*things given). After filling the first box, we now have only (*n*âˆ’1) things to fill the second box; any one of these things can be used to fill the second box and hence the second box can be filled in (*n*âˆ’1) ways; Similarly, the third box can be filled in (*n*âˆ’2) ways and so on the rth box can be filled in (*n*âˆ’(râˆ’1) ways, i.e., [*n*âˆ’r+1] ways. Hence, from the fundamental rules of counting, all the*r*boxes together can be filled up in â†’*n*. (*n*âˆ’1). (*n*âˆ’2)â€¦(*n*âˆ’*r*+1) ways*P*^{n}_{r}_{ }=*n*. (*n*âˆ’1). (*n*âˆ’2)â€¦(*nâ€“r*+ 1)This can be simplified by multiplying and dividing the right hand side be (

*n*âˆ’*r*) (*n*â€“ )â€¦3.2.1.*P*

^{n}*=*

_{r}*n*(

*n*â€“1)(

*n*â€“2)â€¦[

*n*âˆ’ ]

=

The number of arrangements of

*n*distinct things taken*r*things at a time is*P*

^{n}*=*

_{r}If we take

*n*things at a time, then we get*P*^{n}*. From the discussion similar to that we had for filling the*_{n}*r*boxes above we can find that*P*^{n}*is equal to*_{n}*n*! The first box can be filled in*n*ways, the second one in (*n*âˆ’1) ways, the third one in (*n*âˆ’2) ways and so on, then the*nth*box in 1 way; hence, all the*n*boxes can be filled in*P*

^{n}*!*

_{n}= n*r = n*in the formula for

*P*

^{n}*then we get*

_{n}^{ }*P*

^{n}*= ; since we already found that*

_{n}*P*

^{n}*!*

_{n}= nWe can conclude that 0! = 1

The number of combinations of n distinct things taking r at a time =

The number of combinations of n distinct things taking r at a time =

^{n}C_{r}Let the number of combinations

*C*^{n}*be S. Consider one of these S combinations. Since this is a combination, the order of the*_{r}*r*things is not important. If we now impose the condition that order is required for these*r*things, we can get*r*! arrangement from this one combination. So each combination can give rise to*r*! permutations. S combinations will thus give rise to S Ã— r! permutations. But since these are all permutations of*n*things taking*r*at a time, this must be equal to*P*^{n}*. So,*_{r}S Ã—

*r*! =*P*^{n}_{r}_{ }=So, S =

*C*^{n}*= Ã—*_{r}It can also be deduced from here that the number of selections of n distinct things taken all at a time will be equal to 1 (since there is only one way in which all the articles can be selected).

Alternatively

*C*^{n}*= = 1*_{n}^{ }Out of

*n*things kept in a bag, if we select*r*things and remove them from the bag, we are left with (*n*âˆ’1) things inside the bag i.e., whenever*r*things are selected out of*n*things, we automatically have another selection of (*n*âˆ’1) things. Hence, the number of ways of making combinations taking*r*out of*n*things is the same as selecting (*n*âˆ’r) things out of*n*given things, i.e.,*C*

^{n}*=*

_{r}*C*

^{n}

_{n-r}Before we move ahead, let us once again make it clear that whenever we are using

^{n}C_{r}and^{n}P_{r}, our assumption is that all the things are distinct, i.e., no two of them are same.Example-1

Munchun has 10 children. She takes 3 of them to the zoo at a time, as often as she can, but she does not take the same three children to the 0 more than once. How many times Munchun will be required to go to the zoo?

- 120
- 45
- 90
- 180

Solution

Number of times (read ways) 3 children (read distinct things) can be selected from 10 children (read distinct things) =

^{10}C_{3}.So, she will be required to go to the zoo

^{10}C_{3 }times.So, option (a) is the answer.

Example-2

In the above question, how many times a particular child will go?

- 72
- 45
- 90
- 36

Solution

Consider the case for any particular child C

_{1}Since C_{1}has already been selected, so out of the rest 9 children Munchun will be required to select 2 more children. This can be done on^{9}C_{2 }ways.So, option (d) is the answer.

Example-3

In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is

- 200
- 216
- 235
- 256

Solution

Let there be

*m*boys and*n*girls*C*

^{n}_{2}= 45 = â‡’

*n*(

*n*âˆ’1) = 90 â‡’

*n*= 10

*C*

^{m}_{2}= 190 = = 190 â‡’

*m*(

*m*âˆ’ 1) = 380 â‡’

*m*= 20

Number of games between one boy and one girl =

^{10}C_{1}Ã—^{20}C_{1}= 10 Ã— 20 = 200Hence, option (a) is the answer.

Example-4

In how many ways can three persons be seated on five chairs?

Solution

This question is a very fundamental problem of arrangements without repetition. The first person can sit in 5 ways (into any of the five chairs), the second person can take place in 4 ways (into any of the remaining 4 chairs) and the third person can sit in 3 ways.

So, the total number of ways in which these 3 persons can arrange themselves on 5 chairs is 5 Ã— 4 Ã— 3 = 60.