How many different ways can the letters of the word American be arranged so that the vowels are at the two ends?

Permutation is known as the process of organizing the group, body, or numbers in order, selecting the body or numbers from the set, is known as combinations in such a way that the order of the number does not matter.

In mathematics, permutation is also known as the process of organizing a group in which all the members of a group are arranged into some sequence or order. The process of permuting is known as the repositioning of its components if the group is already arranged. Permutations take place, in almost every area of mathematics. They mostly appear when different commands on certain limited sets are considered.

Permutation Formula

In permutation r things are picked from a group of n things without any replacement. In this order of picking matter.

nPr = (n!)/(n – r)!

Here,

n = group size, the total number of things in the group

r = subset size, the number of things to be selected from the group

Combination

A combination is a function of selecting the number from a set, such that (not like permutation) the order of choice doesn’t matter. In smaller cases, it is conceivable to count the number of combinations. The combination is known as the merging of n things taken k at a time without repetition. In combination, the order doesn’t matter you can select the items in any order. To those combinations in which re-occurrence is allowed, the terms k-selection or k-combination with replication are frequently used.

Combination Formula

In combination r things are picked from a set of n things and where the order of picking does not matter.

nCr = n!⁄((n-r)! r!)

Here,

n = Number of items in set

r = Number of things picked from the group

Solution:

Vowels are: I,I,O,E

If all the vowels must come together then treat all the vowels as one super letter, next note the letter ‘S’ repeats so we’d use

7!/2! = 2520 

Now count the ways the vowels in the super letter can be arranged, since there are 4 and 1 2-letter(I’i) repeat the super letter of vowels would be arranged in 12 ways i.e., (4!/2!)

= (7!/2! × 4!/2!) 

= 2520(12)

= 30240 ways

Similar Questions

Question 1: In how many ways can the letters be arranged so that all the vowels came together word is CORPORATION?

Solution:

Vowels are :- O,O,A,I,O

If all the vowels must come together then treat all the vowels as one super letter, next note the R’r letter repeat so we’d use

7!/2! = 2520

Now count the ways the vowels in the super letter can be arranged, since there are 5 and 1 3-letter repeat the super letter of vowels would be arranged in 20 ways i.e., (5!/3!)

= (7!/2! × 5!/3!)

= 2520(20)

= 50400 ways

Question 2: In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged such that the vowels must always come together?

Solution:

Vowels are :- A,A,E,I

Next, treat the block of vowels like a single letter, let’s just say V for vowel. So then we have MTHMTCSV – 8 letters, but 2 M’s and 2 T’s. So there are

8!/2!2! = 10,080

Now count the ways the vowels letter can be arranged, since there are 4 and 1 2-letter repeat the super letter of vowels would be arranged in 12 ways i.e., (4!/2!)

= (8!/2!2! × 4!/2!)

= 10,080(12)

= 120,960 ways

Question 3: In How many ways the letters of the word RAINBOW be arranged in which vowels are never together?

Solution:

Vowels are :- A, I, O  

Consonants are:- R, N, B, W.

Arrange all the vowels in between the consonants so that they can not be together. There are 5 total places between the consonants. So, vowels can be organize in 5P3 ways and the four consonants can be organize in 4! ways.

Therefore, the total arrangements are 5P3 * 4! = 60 * 24 = 1440

Article Tags :

Example 1: How many four-digit numbers can be formed using the digits 0, 3, 4, 5, 6, 7 if

  1. Repetition of digits is not allowed?
  2. Repetition of digits is allowed?

Sol: (i) In a four digit number, 0 cannot appear in the thousand’s place. So, the thousand’s place can be filled in 5 ways (viz. 3, 4, 5, 6, and 7). Since, the repetition of digits is not allowed and 0 can be used at hundred’s place, so the hundred’s place can be filled in 5 ways. Now, any one of the remaining four digits can be used to fill up ten’s place. So, ten’s place can be filled in 4 ways. One’s place can be filled with the remaining three digits in 3 ways. Hence, the required number of ways = 5 × 5 × 4 × 3 = 300.

(ii) For a four digit number, we have to fill up four places and 0 cannot appear in the thousand’s place. So, thousand’s place can be filled in 5 ways. Since, repetition of digits is allowed, so each of the three remaining places viz hundred’s, ten’s and one’s place can be filled in 6 ways. Hence, the required number of ways = 5 × 6 × 6 × 6 = 1,080.

How many different ways can the letters of the word American be arranged so that the vowels are at the two ends?

Example 2: It is required to seat 5 Indians and 4 Americans in a row so that all Americans occupy the even places. How many such arrangements are possible?

Sol: In all, 9 persons are to be seated in a row and in the row of 9 positions; there are exactly four even places viz. second, fourth, sixth and eighth. It is given that these four even places are to be occupied by 4 Americans. This can be done in 4P4  ways. The remaining five positions can be filled by the 5 Indians in 5P5 ways. So, by the fundamental principle of counting, the number of seating arrangements as required is 4P4 × 5P5 = 4! × 5! = 24 × 120 = 2,880.

Example 3: Find the sum of all the numbers that can be formed with the digits 2, 3, 7, 8 taken all at a time.

Sol: The total number of numbers formed with the digits 2, 3, 7 and 8 taken all at a time = Number of arrangements of 4 digits taken = 4P4 = 4! = 24. To find the sum of these 24 numbers we will find the sum of digits at units, tens, hundred’s and thousand’s place in all these numbers. Consider the digits in the unit’s place in all these numbers. Each of the digits 2, 3, 7 and 8 occur in 3! = 6 times in the unit’s place.

So, the total ways for the digits in the unit’s place in all these numbers = (2 + 3 + 7 + 8) x 3! = 120.  Similarly, the sum of the digits in the ten’s, hundred’s and thousand’s places in all these numbers = (2 + 3 + 7 + 8) × 3! = 120 each. Hence the sum of all the numbers = (100 + 101 + 102 + 103) × 120 = 133320.

Example 4: How many words can be formed from the letters of the word ‘HALFTIME’ so that the vowels never come together?

Sol: The total number of words formed by using all the eight letters of the word ‘HALFTIME’ is 8P8 = 8! = 40,320. Now we will find the words in which the vowels are together. There are three vowels A, I and E. Let’s take them as one unit. So we have 5 letters and one unit of vowels. These 6 can be arranged in 6! ways. Also, the 3 vowels can be arranged in 3! ways. So the total number of words in which the vowels are together = 6! × 3! = 720 × 6 = 4320. So, the total number of words in which vowels are never together = Total number of words – Number of words in which vowels are together = 40,320 – 4,320 = 36,000.

Example 5: How many four digit numbers divisible by 4 can be made with the digits 1, 2, 7, 4, 9 if the repetition of digits is not allowed?

Sol:    A number is divisible by 4 if the number formed by the last two digits is divisible by 4.

So, there are four two-digit numbers divisible by 4, which can be made with the help of these digits. These are 12, 24, 72 and 92.

Now corresponding each such way the remaining three digits at thousand’s and hundred’s places can be arranged in 3P2 ways.

Hence the required number of numbers = 3P2 × 4 = 3! × 4 = 24.

Example 6: How many words can be formed using the letter X four times, the letter Y twice and the letter Z twice?

Sol: We are given 8 letters viz. X, X, X, X, Y, Y, Z, Z. Clearly, there are 8 letters of which four are of one kind, two are of second kind and two are of third kind. So, the total number of permutations is (8!/4!2!2!) =420. Hence the required number of words = 420.

Example 7: How many teams of 4 persons can be formed out of 7 men, 3 women and 5 boys if each team has a man and contains at least one woman?

Sol: The following cases can be made to form a team of four persons :

Hence the number of teams = 210 + 105 + 7 = 322

Example 8: Find the number of ways in which 12 different flowers can be arranged to form a garland.

Sol: 12 different flowers can be arranged in circular form in (12 – 1)! = 11! ways. Since there is no distinction between the clockwise and anticlockwise arrangements, so, the required number of arrangements = (11!/2)

Example 9: Out of 5 boys and 2 girls, a committee of 3 is to be formed. In how many ways can it be done if at least one girl is to be included?

Sol: The committee can be constituted in the following ways:

  1. By selecting 2 boys and 1 girl.
  2. By selecting 1 boy and 2 girls

Combination Calculator: 2 boys out of 5 boys and 1 girl out of 2 girls can be chosen in 5C2 × 2C1 ways and 1 boy out of 5 boys and 2 girls out of 2 girls can be chosen in 5C1 × 2C2 ways. ∴The total number of ways of forming the committee = 5C2 × 2C1 + 5C1 × 2C2 = 20 + 5 = 25.

How many different ways can the letters of the word American be arranged so that the vowels are at the two ends?

Example 10:    How many different 11 letter words can be formed with the letters ppppeeeeuuk?

Sol: There are 11 letters in the given word of which 4 are p’s, 4 are e’s and 2 are u’s. The total number of words is the arrangement of 11 things, of which 4 are alike of one kind, 4 are alike of second kind and 2 are of third kind i.e.(11!/4!4!2!) . Hence, the total number of words =(11!/4!4!2!) = 34,650.