Solution : The word OMEGA consists of 5 different letters. Out of which O,E, and A are vowels and M,G are consonants .
[i] Out of 5 places 3 are odd positions `[1^[st] , 3^[rd] and 5^[th]]` and 2 are even places [`2^[nd] and 4^[th]`] Three vowels can be arranged in the odd places in `3!` ways and the two consonants can be arranged in the even place in 2! ways.
`therefore` Number of such words `=3!xx2!""=12`
[ii] Out of the
three vowels [A,E,O] first letter can be arranged in `""^[3]P_[1]` =3 ways. After arranging the first letter remaining 4 positions can be arranged in 4! ways.
`therefore ` Required number of arrangements `=3xx4!""=72`
Permutation:
A Permutation is a collection or a combination of objects from a set where the order or the arrangement of the chosen objects does matter. In other words, a Permutation is an arrangement of objects in a definite order, For example, if we have two elements A and B, then there are two possible arrangements, [ A B ] and [ B A].
Key Point
- nPn = n[n – 1] [n -2]… 3x2x1=n!
- nP0 = 1
- nP1 = n
- nPn-1 = n!
- nPr = n.n-1Pr-1 = n[n-1]n-2Pr-2
The number of permutations when ‘r‘ elements are arranged out of a total of ‘n’ elements is n Pr = n! / [n – r]! For example, let n = 4 [A, B, C and D] and r = 2 [All permutations of size 2]. The answer is 4!/[4-2]! = 12. The twelve permutations are AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, and DC.
Combination:
A Combination is the different selections of a given number of elements taken one by one, or some, or all at a time. For example, if we have two elements A and B, then there is only one way select two items, we select both of them.
Key Point
- nCr is a natural number
- nC0=[nCn]=1
- nC1=n
- nCr=[nCn−r]
- nCx=nCy ⇒x=y or x+y=n
- n.n−1Cr−1=[n−r+1]×nCr−1
The number of combinations when ‘r’ elements are selected out of a total of ‘n’ elements is n C r = n! / [[r !] x [n – r]!]. For example, let n = 4 [A, B, C and D] and r = 2 [All combinations of size 2]. The answer is 4!/[[4-2]!*2!] = 6. The six combinations are AB, AC, AD, BC, BD, and CD.
Note: In the same example, we have different cases for permutation and combination. For permutation, AB and BA are two different things but for selection, AB and BA are the same.
Sample Problems
Question 1: How many words can be formed by using 3 letters from the word “DELHI”?
Solution: The word “DELHI” has 5 different words. Therefore, required
number of words = 5 P 3 = 5! / [5 – 3]!
Required number of words = 5! / 2! = 120 / 2 = 60
Question 2: How many words can be formed by using the letters from the word “DRIVER” such that all the vowels are always together?
Solution: In these types of questions, we assume all the vowels to be a single character, i.e., “IE” is a single character. So, now we have 5 characters in the word, namely, D, R, V, R, and IE. But, R occurs 2 times. => Number of possible arrangements = 5! / 2! = 60 Now, the two vowels can be arranged in 2! = 2 ways. => Total number of possible words such that the vowels are always together= 60 x 2 = 120
Question 3: In how many ways, can we select a team of 4 students from a given choice of 15?
Solution : Number of possible ways of selection = 15 C 4 = 15 ! / [[4 !] x [11 !]]
Number of possible ways of selection = [15 x 14 x 13 x 12] / [4 x 3 x 2 x 1] = 1365
Question 4: In how many ways can a group of 5 members be formed by selecting 3 boys out of 6 boys and 2 girls out of 5 girls?
Solution : Number of ways 3 boys can be selected out of 6 = 6 C 3 = 6 ! / [[3 !] x [3 !]] = [6 x 5 x 4] / [3 x 2 x 1] = 20 Number of ways 2 girls can be selected out of 5 = 5 C 2 = 5 ! / [[2 !] x [3 !]] = [5 x 4] / [2 x 1] = 10 Therefore, total number of ways of forming the group = 20 x 10 = 200
Question 5: How many words can be formed by using the letters from the word “DRIVER” such that all the vowels are never together?
Solution: we assume all the vowels to be a single character, i.e., “IE” is a single character. So, now we have 5 characters in the word, namely, D, R, V, R, and IE. But, R occurs 2 times. => Number of possible arrangements = 5! / 2! = 60 Now,
the two vowels can be arranged in 2! = 2 ways. => Total number of possible words such that the vowels are always together = 60 x 2 = 120 ,
total number of possible words = 6! / 2! = 720 / 2 = 360 Therefore, total number of possible words such that the vowels are never together 240