Permutation :- It can be defined as the number of ways in which r things at a time can be SELECTED and ARRANGED at a time from amongst n things. It is represented by $^{n}\textrm{P}_{r}$.
$^{n}\textrm{P}_{r}$ = Number of permutations (arrangements) of n things taken r at a time.
$^{n}\textrm{P}_{r}$ = $\frac{n!}{(n-r)!}\; ;\; n\geq r!$
The difference between selection and arrangement can be seen through the illustration given below:-
Selection :- Suppose we have three men A, B and C. Out of which two men have to be selected for two posts.
This can be done in the following ways: AB, AC or BC. Physically, they can be counted as three distinct selections. This value can also be got by using $^{3}\textrm{C}_{2}$.
Note:- Here, we are counting AB and BA as one single selection and same for AC, CA, BC and CB.
Arrangement :- Suppose we three men A, B and C. Out of which two men have to be selected to the post of a captain and vice-captain of a team.
In this case, we have to take AB and BA as two different instances since the order of the arrangement makes a different in who is the captain and vice-captain?
Similarly, we have BC, CB, AC and CA as four more instances. Thus, in all three could be six arrangements of two things out of three.
This is given by $^{3}\textrm{P}_{2}$ = 6
Combinations :- It can be defined as the number of ways in which r things at a time can be selected from amongest n things available for selection. It is represented by $^{n}\textrm{C}_{r}$
$^{n}\textrm{C}_{r}$ = Number of combinations (selection) of n things taken r at a time.
$^{n}\textrm{C}_{r}$ = $\frac{n!}{r!(n-r)!}\; ;\; n\geq r$
Relation between Permutation and Combination :-
When we look at the formulae for permutations and combinations and compare the two, we see that
$^{n}\textrm{P}_{r}=r!\times ^{n}\textrm{C}_{r}$
= $^{n}\textrm{C}_{r}\times \; ^{r}\textrm{R}_{r}$
MNP Rule :- If there are three things to do and there are M ways of doing the first thing, N ways of doing the second thing and P ways of doing the third thing, then there will be $M\times N\times P$ ways of doing all the three things together.
For eg. The numbers 1, 2, 3, 4 and 5 are to be used for forming 3 digit numbers without repetition. In how many ways can this be done?
Using the MNP rule we can visualize this as -
There are three things to do -
First digit can be selected in 5 distinct ways,
Second digit can be selected in 4 distinct ways, and
Third can be selected in 3 different ways.
$\therefore $ Total number of 3 digit numbers can be formed are $5\times 4\times 3=60$.
$^{n}\textrm{P}_{r}$ = Number of permutations (arrangements) of n things taken r at a time.
$^{n}\textrm{P}_{r}$ = $\frac{n!}{(n-r)!}\; ;\; n\geq r!$
The difference between selection and arrangement can be seen through the illustration given below:-
Selection :- Suppose we have three men A, B and C. Out of which two men have to be selected for two posts.
This can be done in the following ways: AB, AC or BC. Physically, they can be counted as three distinct selections. This value can also be got by using $^{3}\textrm{C}_{2}$.
Note:- Here, we are counting AB and BA as one single selection and same for AC, CA, BC and CB.
Arrangement :- Suppose we three men A, B and C. Out of which two men have to be selected to the post of a captain and vice-captain of a team.
In this case, we have to take AB and BA as two different instances since the order of the arrangement makes a different in who is the captain and vice-captain?
Similarly, we have BC, CB, AC and CA as four more instances. Thus, in all three could be six arrangements of two things out of three.
This is given by $^{3}\textrm{P}_{2}$ = 6
Combinations :- It can be defined as the number of ways in which r things at a time can be selected from amongest n things available for selection. It is represented by $^{n}\textrm{C}_{r}$
$^{n}\textrm{C}_{r}$ = Number of combinations (selection) of n things taken r at a time.
$^{n}\textrm{C}_{r}$ = $\frac{n!}{r!(n-r)!}\; ;\; n\geq r$
Relation between Permutation and Combination :-
When we look at the formulae for permutations and combinations and compare the two, we see that
$^{n}\textrm{P}_{r}=r!\times ^{n}\textrm{C}_{r}$
= $^{n}\textrm{C}_{r}\times \; ^{r}\textrm{R}_{r}$
MNP Rule :- If there are three things to do and there are M ways of doing the first thing, N ways of doing the second thing and P ways of doing the third thing, then there will be $M\times N\times P$ ways of doing all the three things together.
For eg. The numbers 1, 2, 3, 4 and 5 are to be used for forming 3 digit numbers without repetition. In how many ways can this be done?
Using the MNP rule we can visualize this as -
There are three things to do -
First digit can be selected in 5 distinct ways,
Second digit can be selected in 4 distinct ways, and
Third can be selected in 3 different ways.
$\therefore $ Total number of 3 digit numbers can be formed are $5\times 4\times 3=60$.
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