Maths > Permutations and Combinations > 2.0 Basic Principle of Counting
Permutations and Combinations
1.0 Factorial notation
2.0 Basic Principle of Counting
3.0 Permutation
4.0 Circular Permutations
5.0 Combinations
6.0 Restricted Selection
7.0 Restricted Arrangement
8.0 Greatest Term
9.0 Possible selections from $n$ distinct objects
10.0 Possible Selection from $n$ Identical Objects
11.0 Possible Selection from $n$ Objects having Distinct and Identical Objects
12.0 Total Number of Possible Divisors for a Given Natural Number
13.0 Sum of all Possible Divisors of a Natural Number
14.0 Exponent of a Prime Number in $n!$
15.0 Division and Distribution of Objects
16.0 Multinomial Theorem
2.2 Rule of Sum
Statement: If one experiment has '$n$' possible outcomes and another experiment has '$m$' possible outcomes, then there are '$m + n$' possible outcomes when exactly one of these experiments is performed.
Proof: Let there be two events say $M$ and $N$ ,
Let the outcomes of $M$ be, $${a_1},{a_2},........,{a_m}$$
and the outcomes of $N$ be, $${b_1},{b_2},........,{b_n}$$
Then one can choose the following way,
$${a_1},{a_2},........,{a_m}\quad {\text{and}}\quad {b_1},{b_2},........,{b_n}$$
Here there are '$m$' elements of $M$ and '$n$' elements of $N$ .
Thus the total number of elements is '$m + n$'.
Question 3. In a juice shop there are two categories namely milkshakes and ice-creams. There are $5$ flavours of milkshake and $7$ flavours of ice-cream. Find the possible number of choices you have in case you wish to have only one of these items.
Solution: You can choose one from $5$ milkshakes or from $7$ ice-creams.
Thus the total number of choices one has to choose exactly one item is, $$=5 + 7 = 12$$
Note: The sum rule can be applied for any number of events occurring together.
Question 4. For a class committee meeting in a school, the pupil leader has to choose one from each class. Class $10$ has three sections. The section $A$ has $5$ representatives, section $B$ has $3$ and section $C$ has $4$ for the meeting, from which the pupil leader has to select only one. Find the number of possible choices he has in choosing a class $10$ student for the committee.
Solution: The pupil leader can choose that one student from the $5$ students of section $A$ or from the $3$ students of section $B$ or from the $4$ students of section $C$.
Thus the total number of choices is, $$= 5 + 3 + 4 = 12$$
Question 5. A password has to be set for two firms, Firm $A$ and Firm $B$ with the below conditions. Find the maximum number of unique passwords that can be formed for each of the firms.
| Firm $A$ | Firm $B$ | |
| Number of digits | $9$ digit password | $8$ digit password |
| Digits to be used | $1$ to $9$ | $2$ to $9$ |
| Condition | Without repetition of digits in password | Repetition allowed |
Solution:
Firm $A$: Let each box represent one digit in the password.
| $9$ | $8$ | $7$ | $6$ | $5$ | $4$ | $3$ | $2$ | $1$ |
There are $9$ ways of filling the last box, eight ways of filling next as repetition is not allowed and so on. Thus maximum possible passwords that can be formed is, $$ = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 9! = 3,62,880$$
Firm $B$: Let each box represent one digit in a password.
| $8$ | $8$ | $8$ | $8$ | $8$ | $8$ | $8$ | $8$ |
There are $8$ ways of filling the last box, $8$ ways of filling next as repetition is allowed and so on.
Thus maximum possible passwords that can be formed is, $$ = 8 \times 8 \times 8 \times 8 \times 8 \times 8 \times 8 \times 8 = {8^8} = 1,67,772,16$$
Question 6. There are two groups of circles $A$ and $B$. Group $A$ consists of $3$ circles ${A_1}$ , ${A_2}$ and ${A_3}$, having different centers and in such a way that ${A_2}$ lies entirely in ${A_1}$, and ${A_3}$ lies entirely in ${A_2}$. Group $B$ consists of $4$ circles ${B_1}$ , ${B_2}$, ${B_3}$ and ${B_4}$, having different centers and in a similar fashion. The two groups of circle are in such a way that every circle of group $A$ intersects with every circle of group $B$, as shown in figure.
$\left( i \right)$ How many centers the circles altogether have?
$\left( ii \right)$ How many common chords are obtained?
Solution: $\left( i \right)$ There are $3$ centers in the first and $4$ in the second group. Thus $3+4$ circles are there.
$\left( ii \right)$ To obtain a common chord one circle from each group has to be chosen.
Thus applying product rule, it it $3 \times 4$ i.e. $12$ common chords.
