Fundamental Principle of Counting:
If any event can occur in m ways and after it happens in any one of these ways, a second event can occur in n ways, then both the events together can occur in m \times n ways.
Example 1:
How many multiples of 5 are there from 10 to 95?
Solution:
The first digit from the right can be chosen in 2 ways.
The second digit can be any one of 1, 2, 3, 4, 5, 6, 7, 8, 91, 2, 3, 4, 5, 6, 7, 8, 9
i.e. There are 9 choices for the second digit.
Thus, there are 2 Ã— 9 = 2 Ã— 9 = 18 multiples of 5 from 10 to 95.
Example 2:
In a city, the bus route numbers consist of a natural number less than 100, followed by one of the letters A,B,C,D,E and F. How many different bus routes are possible?
Solution:
There are 99 choices for the number.
The letter can be chosen in 6 ways.
Number of possible bus routes are 99 Ã— 6 = 99 Ã— 6 = 594
Example 3:
There are 3 questions in a question paper. If the questions have 4,3 and 2 solutions respectively, find the total number of solutions.
Solution:
=> By the multiplication (counting) rule,
Total number of solutions=4Ã—3Ã—2==4Ã—3Ã—2= 24
Example:
Suppose you want to arrange your English, Hindi, Mathematics, History, Geography and Science books on a shelf. In how many ways can you do it?
Solution:
The number of permutations of n objects is \(n!\) = \(n.(nâˆ’1).(nâˆ’2)…2.1\)
Here \(n\) = 6 and therefore, number of permutations is 6.5.4.3.2.1 = 720.
The number of permutations of \(r\) objects out of \(n\) objects is
The number of permutations of \(r\) objects out of \(n\) objects is usually denoted by \(^{n}P_{r}\).
Example:
If you have 6 New Year greeting cards and you want to send them to 4 of your friends, in how many ways can this be done?
Solution:
This number is \(^{6}P_{4}\) = 6(6âˆ’1)(6âˆ’2)(6âˆ’3) = 6 x 5 x 4 x 3 = 360
Therefore, cards can be sent in 360 ways.
So, using the factorial notation, this formula can be written as follows:
\(^{n}P_{r}\) = \(\frac{n!}{(nâˆ’r)!}\)
Permutations under Some Conditions:
Example 1:
Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?
Solution:
Then, Parvin cannot be allotted bed number 2.
So Parvin can be allotted a bed in 5 ways.
After allotting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.
So, in this case the beds can be allotted in 5 Ã— 5! = 600 ways.
Case 2:
Anju is allotted bed number 7.
Then, Parvin cannot be allotted bed number 6
As in Case 1, the beds can be allotted in 600 ways.
Case 3:
Anju is allotted one of the beds numbered 2, 3, 4, 5 or 6
Parvin cannot be allotted the beds on the right hand side and left hand side of Anju’s bed.
For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.
Therefore, Parvin can be allotted a bed in 4 ways in all these cases.
After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.
Therefore, in each of these cases, the beds can be allotted 4 Ã— 5! = 480 ways.
=> The beds can be allotted in:
2 Ã— 600 + 5 Ã— 480 = 1200 + 2400 = 3600 ways
Example 2:
In how many ways can an animal trainer arrange 5 lions and 4 tigers in a row so that no two lions are together?
Solution:
| L | T | L | T | L | T | L | T | L |
The 5 lions should be arranged in the 5 places marked ‘L’.
This can be done in 5! ways.
The 4 tigers should be in the 4 places marked ‘T’.
This can be done in 4! ways.
Therefore, the lions and the tigers can be arranged in 5! Ã— 4! = 2880 ways
Example 3:
How many arrangements of the letters of the word â€˜BENGALIâ€™ can be made
(ii) If the vowels are to occupy only odd places.
Solution:
(i) Considering vowels a, e, i as one letter, we can arrange 4 + 1 letters in 5! ways in each of which vowels are together. These 3 vowels can be arranged among themselves in 3! ways.
=> Total number of words = 5! Ã— 3!
= 120 Ã— 6 = 720
So there are total of 720 ways in which vowels are ALWAYS TOGEGHER.
Now,
Since there are no repeated letters, the total number of ways in which the letters of the word â€˜BENGALIâ€™ cab be arranged:
= 7! = 5040
So,
Total no. of arrangements in which vowels are never together:
= ALL the arrangements possible – arrangements in which vowels are ALWAYS TOGETHER
= 5040 âˆ’ 720 = 4320
(ii) There are 4 odd places and 3 even places. 3 vowels can occupy 4 odd places in \(^{4}P_{3}\) ways and 4 constants can be arranged in \(^{4}P_{4}\) ways.
=> Number of words = \(^{4}P_{3}\) x \(^{4}P_{4}\) = 576.
Combination:
Each of the different groups or selections which are formed by taking some or all of a number of things irrespective of order is called a combination.
Example: All the combinations formed by a, b, c taking two at a time are ab, bc, ca.
Note that ab and ba are two different permutations but represent the same combination.
Example 1:
Find the number of subsets of the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} having 4 elements.
Solution:
We have to find the number of ways of choosing 4 elements of this set which has 11 elements.
This can be done in:
\(^{11}C_{4}\) = \(\frac{11 \times 10 \times 9 \times 8}{1 \times 2 \times 3 \times 4}\) = 330 ways.
Example 2:
12 points lie on a circle. How many cyclic quadrilaterals can be drawn by using these points?
Solution:
Therefore, we can draw 495 quadrilaterals.
Example 3:
The Indian Cricket team consists of 16 players. It includes 2 wicket keepers and 5 bowlers. In how many ways can a cricket eleven be selected if we have to select 1 wicket keeper and at least 4 bowlers?
Solution:
or, 1 wicket keeper and 5 bowlers.
Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players
= \(^{2}C_{1}\) X \(^{5}C_{4}\) X \(^{9}C_{6}\) = 840
Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players
= \(^{2}C_{1}\) X \(^{5}C_{5}\) X \(^{9}C_{5}\) = 252
=> Total number of ways of selecting the team:
= 840 + 252 = 1092
Definition of rank:
If the words, that can be formed from the letters of a given word, are listed as in a dictionary and if given word appears at the nth place in this list, then n is called the rank of the given word(in the list).
2. If there are n objects of which \(p_{1}\) are alike of one kind; \(p_{2}\) are alike of another kind; \(p_{3}\) are alike of third kind and so on and \(p_{r}\) are alike of rth kind, such that (\(p_{1} + p_{2} + p_{3 + …… + p_{r}}\)) = n. Then,
number of permutations of these n objects = \(\frac{n!}{(p_{1}!)(p_{2}!)……p_{r}!}\)
3. The number of all combinations of n things, taken r at a time is given by:
\(n_{{c}_{r}}\) = \(\frac{n!}{(r!)(n – r)!}\) = \(\frac{n(n – 1)(n – 2)….(n – r + 1)}{1.2.3….(r – 1).r}\)
4. \(n_{{c}_{r}}\) =\(n_{{c}_{n – r}}\)
5. If \(n_{{c}_{r}}\) = \(n_{{c}_{s}}\) then either r = s or r + s = n
6. \(n_{{p}_{r}}\) = n – \(1_{{p}_{r}} + r * n – 1_{{p}_{r – 1}}\)
7. \(n_{{c}_{r}}\) = n – \(1_{{c}_{r – 1}} + n – 1_{{c}_{r}}\)
8. \(\frac{n_{{p}_{r}}}{n – 1_{{p}_{r – 1}}}\) = n and \(\frac{n_{{c}_{r}}}{n – 1_{{c}_{r – 1}}}\) = \(\frac{n}{r}\)
9. \(\frac{n_{{p}_{r}}}{n_{{p}_{r – 1}}}\) = n – r + 1 and \(\frac{n_{{c}_{r}}}{n_{{c}_{r – 1}}}\) = \(\frac{n – r + 1}{r}\)
2. In how many ways can a cricket eleven be chosen out of a batch of 15 players?
Solution:
3. How many words can be formed by using all the letters of the word ‘SUCCESS’, so that the vowels are never together?
Solution:
4. Compute: \(\frac{35!}{30!}\)?
Solution:
5. Find the value of:
Solution: