Ram wants to go to a trip with his friends, so he has packed three shirts and two pants with him. He thinks that if he has put on a different shirt and a different pant, then a new pair is formed and he can wear different pairs on different days. Now the question arises is that how many days he can wear different pairs? This question simply means that we have to tell the no. of pairs which can be formed from three shirts and two pants.
The answer is simple, that is 6.
This is because for each shirt, we have two options of pants, so for three shirts we have six pair.
Did you know that this is nothing but fundamental principle of counting!!
Now let’s see the definition of this principle according to ncert-:
Suppose an event E can occur in m different ways and associated with each way of
occurring of E, another event F can occur in n different ways, then the total number of
occurrence of the two events in the given order is m × n
Now let’s suppose there are five given no. s as 1,2,3,4,5 and we have to make a three digit no. using these no.s only but we can’t repeat any no. Basically, we have to fill up three dashes ____ ____ ____ by picking up any no. from the given and to find how many patterns can be made. Now, we can choose any no. for the first place, means we have five options to fill the first place (from LHS), now once we have chosen a given no. for the first place, we have four options for the second place(as repetition of no.s is not allowed). Now after this, we have three options for the last place. So, total no. of patterns that we can make is 5 × 4 ×3.
We just have applied fundamental principle of counting!!
Now let’s talk about permutations.
First let’s see the definition from Google-:
each of several possible ways in which a set or number of things can be ordered or arranged.
So basically 213 is a permutation as it is a possible way of arranging no.s from the given five no. s.
In permutation, order matters!!
Means 213 is different from 231.
Actually, there is a thing where order does not matters that is combination. But we will study this topic later.
Now let’s find out a general formula for calculating the no. of permutations for any problem!
Suppose we have to choose n things from m options and want to make a pattern just like in the above question, then we can find out our answer by drawing n dashes
__ __ __ __……… up to n
And then filling up the dashes starting from m like this-:
m×(m-1) ×(m-2)….. (m-n+1)
Did know that why the last term is (m-n+1)?
For the first term: m =m-1+1
For the second term: m-1=m-2+1
For the nth term:
We denote the above no. of permutations as:
Now, writing the whole statement up to n terms is really a hard work.
So, we must find a better way to illustrate or find the no. of permutations.
For this,we must know the meaning of Factorial.
Factorial of a no. is nothing but the product of that no. and the previous no. and further previous no.s up to 1.
For example, factorial of 5, denoted by 5! is 5×4×3×2×1
Also, 0! =1.
To know why 0!=1, see the video below :
Now, the formula for calculating permutations is somewhat similar to factorial but the difference is just that it doesn’t go up to 1 but stops
at a certain term which is nth term from the beginning.
The last term of permutation formula is: (m-n+1)
The next term supposed to be (m-n).
Let’s multiply and divide (m-n)!
to the formula of permutation.
m×(m-1) ×(m-2)….. (m-n+1) ×(m-n)!
Now, we know that 5×4! is nothing but 5! Similarly 5×4×3! is also 5!
So the numerator can be written as
So our formula for finding the no. of permutations becomes-:
Now let’s solve a question based on permutation and principle of counting:
Q How many three digit even no.s can be made using the digits 1,2,3,4,6,7,if no digit is repeated?
Ans The only thing we must keep in our mind is the property of even no.s that is the last digit of any even no. must be a multiple of 2 like 2,4,6,8.
Basically we have no. of dashes as
3: ___ ___ ___
The last place must be filled with an even no.
Now, from the given no.s we have
2,4,6 as even no.s.
Means the last place can be filled with three no.s only.
After filling up the last place, we are left with 5 no.s to be filled on the second place and then 4 no.s left from which we can choose a no. for the first digit.
So, according to fundamental principle of counting,
Total three digit even no.s are:
4×5×3 which is 60
Q From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?
Ans The answer to this question is simply permutation based. Choosing two from eight gives us:
If repetition is allowed then the answer is simply 8×8 that is 64.