Permutation and combination is one of the fundamental topics in maths that comes with a plethora of applications in modern technology and in its evolution.It is widely used in statistics, probability and in the solution of many real world problems.So let's see how the knowledge of P & C plays an instrumental role in our daily life.
What is Permutation
Whenever we talk of arrangements we can use permutation there Or when you want to know possible arrangements by putting things in a specific order. It would be easier to understand it with simple examples to complex ones.
The formula to arrange r items from given n distinct items is:
\( {}^nP_r \) = \( n! \over { (n-r)!}\)
Example 1 : How many meaningful or meaningless 2 letter words can be made from the word "Hi"?
Let's count the number of letters:
1. Hi
2. iH
3.HH
4. ii
In this word we allowed repetition of letters so we got 4 words but If it was not allowed we would get 2 words only.
1. Hi
2.iH
Example 2 : How many meaningful or meaningless 3 letter words can be formed from the letters of "PIN"
Let's see the possibile arrangements:
1.PIN
2.INP
3.NPI
4.PNI
5.NIP
6.IPN
7.PPP
8.NNN
9.III
Total number of words= 9
Without Repetition the total number of words= 6
Example 3: A digital lock opens with a three digit number.but the owner of the lock forgot third digit and the starting two digits are 4 and 2.Now let's see how many password he can try to open this lock?
Ans : Since two digits are known so we have to ascertain the last digit only.Hence the number of Permutations
4,2,0
4,2,1
4,2,2
4,2,3
4,2,4
4,2,5
4,2,6
4,2,7
4,2,8
4,2,9
So as you can see there is a total of 9 permutations exist.
Or we can simply say that the last digit could be any number from 0 to 9 Hence there are 9 possible passwords to open the lock.
Or The total number of Possible 3 digit passwords =\({}^9P_1\)
= \(\frac{9!}{(9-1)!}\)
= \(\frac{9 \times 8!}{8!}\)
= 9 passwords
= \(\frac{9 \times 8!}{8!}\)
= 9 passwords
What is combination
When the order of things is not important while grouping them we are generally use combination.In combination the order of things don't matter.
The formula to group r items from given n distinct items is :
\( {}^nC_r \) = \( n! \over { r! (n-r)!}\)
Example 1 : Carry,Adam and Neo three friends want to form a group of two friends.how many such groups can be formed?
However permutation can be used but we can't use permutation to get correct result.we will first use permutation then combination.
Let's see the total number of permutation
1. Carry, Adam
2. Adam,Carry
3. Neo, Adam
4. Adam,Neo
5. Carry,Neo
6.Neo, Carry
Using permutation the number of groups= 6.
But practically only 3 groups are possible.
Why three groups?
Let's understand.
Carry,Adam or Adam, carry are not different groups they both points to the same group.as we have discussed earlier,In combination order doesn't matter.
Similarly, The rest of 4 group give 2 unique groups.
Total number of unique groups = 3 groups.
Example 2 : Five friends(A,B,C,D,E) gather for a function and they all shake hand with each other.What are the number of handshakes ?
Ans : There are five friends so 5 hands are used for shaking.For one handshake at least two hands are required.
Total number of Permutations for Handshakes:
1. AB (1 handshake)
2. BA
3. AC
4. CA
5. AD
6. DA
7. AE
8. EA
9. BC
10. CB
11. BD
12. DB
13. BE
14. EB
15. CD
16. DC
17. CE
18. EC
19. DE
20. EB
There are Total 20 permutations for Handshakes. Since the permutations AB and BA point to one handshake and similar logic applies to the other Permutations as well.
Hence,The total number of handshakes in the function will be calculated by unique combination of handshake.
Total unique combinations
So the number of handshakes in the function:
\({}^5C_2\) = \(\frac{5!}{2!(5-2)!}\)
= \(\frac{5!}{2! \, 3!}\)
= \(\frac{5 \times 4 \times 3!}{2! \, 3!}\)
= \(\frac{5 \times 4 \times 3!}{2! \, 3!}\)
=10
Total number of handshakes in the function= 10
Example 3 : Now Let's make a little twist in the previous instance. There are 7 Students joining the party and 3 of them are girls Now the Questions are as follows:
(i) Find the number of handshakes in the party
(ii) Find the number of handshakes that took place among the girls and boys only
(iii) Find sum of handshakes among girls only and among Boys only.
Ans :
(i) Total number of handshakes in the party = \( {}^7C_2 \)
= \(\frac{7!}{2!(7-2)!}\)
= \(\frac{7!}{2! 5!}\)
= \(\frac{7 \times 6 \times 5!}{2! \, 5!}\)
= \(\frac{7 \times 6 \times 5!}{2! \, 5!}\)
= 21 handshakes
(ii) The number of handshakes that took place among Boys and girls only is simply calculated by the product of number of boys and number of girls.
Number of girls = 3
Number of boys= 7-3 = 4
Total number of handshakes among girls and boys only = 4×3 =12 handshakes.
The sum of handshakes among Boys only and among girls only = Total handshakes in the party - Number of handshakes among girls and boys only
= 21 - 12
= 9 handshakes
Alternative Way : This problem can also be solved by the sum of handshakes among Boys only and handshakes among girls only.
= \( {}^4C_2 + {}^3C_2 \)
=\( \frac{4!}{2!(4-2)!} + \frac{3!}{2!(3-2)!} \)
= \( 6 + 3 \)
= 9 handshakes
FAQs Based on Permutations & Combinations
Q 1 : How to decide whether to use permutation or Combination?
Ans : We use permutation where Order is important and we use Combination where Order does not matter, Only grouping matters.
Q 2 : What is Circular Permutation?
Ans : In circular permutation the number of Possible arrangements are observed circularly keeping one element at fixed point. And it is given by
(n -1) !