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# Math Examples of Permutations and Combinations

• ## In how many ways 6 girls & 6 boys sit around in a circle such that no girl is adjacent to each other?

Counting can be done using number of methods like Permutation, combination, factorization etc.
Let us consider an example of round table, which is a very famous question of permutation in mathematics.

Q. In how many ways 6 girls and 6 boys sit around in a Circle such that no girl is adjacent to each other?
Solution: According to problem, repetition of girls and boys in the complete arrangement is not possible because one boy or one girl can acquire one seat at a time. Also there is one more restriction in the problem that no two girls can sit adjacent to each other. This makes it necessary for each girl to be between two boys or adjacent to only one boy. We have to make 6 boys and 6 girls sit in a circle. Number of ways for this arrangement can be calculated as follows:

6 boys

6 girls

5 boys

5 girls

4 boys

4 girls

3 boys

3 girls

2 boys

2 girls

1 boys

1 girls

Arrangement being circular, we can start the sitting arranging from any Position such that we can make either a girl sit first or a boy sit first. Whosoever is made to sit at first, we would be getting the same arrangement as shown above. Here, the first position can be filled in 6 ways i.e. we have to make 1 boy sit out of available 6. Next position has to be filled by a girl according to the restriction that has been imposed on the problem. This place can also be filled in 6 possible ways. Next two places also have to be filled like previous two but by deduction of boy and one girl have to be made. This way we can arrange 6 boys an 6 girls in a circle, getting the total number of ways = 6!2.

## A bag contains 6 white and 4 black balls. two balls are drawn at random. Find the probability that they are of the same color?

Problems related to Probability distribution can be solved either by taking Permutation or Combination of events. For instance, question is:
A bag contains 6 white and 4 black balls two balls are drawn at random, find the probability that they are of the same color?
In such problems, at first we must collect data i.e. events that are present in problem. In our problem we have 6 white balls and 4 black balls in the bag and out of them we have to choose two balls such that they are of same color.

At first we need to decide, whether we should use permutation or combination. In our problem the combinations of different events have to be considered. There can be two possibilities for two balls to be of same color. Either the balls are being picked from 6 white balls or from 4 black balls i.e. they can both be either white or both can be black.
Number of possible cases for two balls that are picked to be white: 6C2
Similarly, number of possible cases for two balls that are picked to be black: 4C2
Total number of balls in bag: 6 + 4 = 10.
Number of ways two balls can be picked from the bag: 10C2. This is our Sample Space.
So, probability of two balls to be of same color can be calculated as:
Or P = (6C2 + 4C2) / 13C2
= ((6 * 5 * 4 * 3 * 2 * 1) / (2 * 4 * 3 * 2 * 1) + (4 * 3 * 2 * 1) / (2 * 2)) / ((13 * 12)/ 2),
= (15 + 6)/78 = 21/78 = 0.27.

## In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

In the given word MISSISSIPPI, I appears 4 times, S appears 4 times, P appears 2 times and M appears once.
Number of distinct permutations of the letters in the given word is

When the four I’s come together, they can be considered as a single letter and this single letter with the remaining 7 letters together accounts to 8 letters.
These 8 letters with 4 S’s and 2 P’s can be arranged in  ways = ways.
Therefore the number of ways where all the 4 I’s  occur together is 840.
The number of  distinct permutations of the letters in MISSISSIPPI where  the four I’s do not come together is 34650 – 840 = 33810.

## If the different permutations of the word EXAMINATION are arranged as in dictionary, how many words are there before the first word starting with E?

In the word EXAMINATION there are 11 letters of which A, I and N appear twice.

The words that will be listed before the words starting with E in the dictionary are those that start with A.
So when the letter A is fixed at the extreme left Position then the arrangement of the remaining 10 letters taken all at a time with 2 I’s and 2 N’s is

The number of words starts with A and that which appears before the letter starting with E in the word EXAMINATION is 907200.

## If all the letters of the word AGAIN be arranged as in dictionary, what is the fiftieth word?

There are 5 letters in the word AGAIN in which A appears twice.
Therefore the required number of words =  = 60
For the words that start form A, we fix the letter A at the extreme left Position and arrange the remaining 4 letters taken all at a time and that is 4P4 =4! = 24
Then for the words that start with G, we fix the letter G at the extreme left position and arrange the remaining 4 letters A, A, I, N taken all at a time
=  = 12
Similarly there are 12 words starting from the letter I.
So the total number of words obtained so far will be 24 + 12 + 12 = 48.
Therefore the 49th word would be NAAGI and the 50th word would be NAAIG.

## How many words each of 3 vowels and 2 consonants can be formed from the letters of the word “INVOLUTE”?

In the word INVOLUTE, there are 4 vowels   (I, O, U, E) and 4 consonant (N, V, L, T).
The number of ways of selecting 3 vowels out of 4 = 4C2
The number of ways of selecting 2 consonants out of 4 = 4C3
Also the 3 vowels and 2 consonants can be arranged in 5 ! ways.
Therefore the required number of words = 4C3 . 4C2 . 5!
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