How many 4 letter words can be formed from the letters of the word combination so that the vowels always come together?

As the name implies, a number system is a mathematical system that is used to represent numerals using various symbols and variables. Under the number system, numbers that can be plotted on a number line, commonly known as real numbers, are represented by a set of values or quantities. Based on their various features, distinct sorts of numbers are classified into different sets or groups. For example, rational numbers are any integers that can be represented in the form p/q, where q is a non-zero integer. Decimal, binary, octal, and hexadecimal are examples of different sorts of systems.

Combinations

It is defined as the process of choosing one, two, or a few elements from a given sequence, regardless of the order in which they appear. If you choose two components from a series that only has two elements to begin with, the order of those elements won’t matter.

Combination Formula

When r items are chosen from n elements in a sequence, the number of combinations is

nCr = n! / r! (n – r)!

For example, let n = 7 and r = 3, then number of ways to select 3 elements out of 7 = 7C3 = 7!/3!(7 – 3)! = 35.

Solution:

If the repetition of letters is allowed, then each alphabet can be chosen from the given 26 alphabets.

Hence total number of combinations = 26C1 × 26C1 × 26C1

= 26 × 26 × 26 

= 17576

However, if repetition is not allowed, then the number of combinations = 26C1 × 25C1 × 24C1 

= 26 × 25 × 24

= 15600

Similar Problems

Problem 1. Given a piggy bank containing 20 coins, determine the number of permutations of nickels, dimes, and quarters it holds.

Solution:

In this case, the order of the coins clearly does not matter. There’s also no mention of whether or not recurrence is permitted.

Following the formula, we have: n+r−1Cr = n+r−1Cn−1 , in the case of r number of combinations from a sequence with n number of elements where elements can be repeated:

The number of possible pairings = 20+3-1C20 = 22C20

= 22! / (22 – 20)! 20!

= 11(21)

= 231

Problem 2. Tell me how many different methods there are to allocate 7 students to a college trip if we only have one triple room and two double rooms.

Solution:

This problem might be understood as having to divide the seven pupils into three groups of three, two, and two students.

Number of ways to choose 3 students in the triple = 7C3 = 7! / 3!4! = 35  

Number of ways to choose 2 out of the remaining 4 students = 4C2 = 4!/ 2!2! = 6

The number of possible ways to choose two students from the remaining two students is equal to one.

Total number of arrangements = 35 × 6 × 1 = 210.

During a conference, 7 students can be assigned to 1 triple and 2 double hotel rooms in 210 different ways.

Problem 3. Determine the number of ways a five-person committee may be established from a group of seven men and six women, with at least three men on the committee.

Solution:

At least three men on the committee means we can have either exactly three, four or all five men in the committee.

Number of arrangements when there are 3 men and 2 women on the committee = (7C3 x 6C2) = 525

Number of arrangements when there are 4 men and 1 woman on the committee=  (7C4 x 6C1) = 210

Number of arrangements when there are all 5 men on the committee = (7C5) = 21

Total arrangements = 525 + 210 + 21

= 756

Problem 4. Find the number of ways the letters in the word ‘LEADING’ can be arranged so that the vowels always appear together.

Solution:

If the vowels are to appear together, they would form a separate letter in the word. Hence we are left with 4 + 1 = 5 letters, which can be arranged in 5! = 120 ways.

Furthermore, there are 3! = 6 ways to arrange the vowels together.

Total number of ways of arranging the letters = 120 x 6 = 720.

Problem 5. Find the number of words with four consonants and three vowels that may be made from eight consonants and five vowels.

Solution:

Number of ways of selecting 4 consonants out of 8 and 3 vowels out of 5 = 8C4 x 5C3

= \frac{8 ×7 ×6 ×5 ×4!}{4!  × 4!}  × \frac{5 ×4 ×3!}{3!  × 2!}

= 70 × 10 = 700

Number of ways of arranging the 7 letters among themselves = 7! = 5040

Number of words that can be formed = 5040 × 700 = 3528000.

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10 Questions 10 Marks 10 Mins

Concept:

If n is a positive integer and r is a whole number, such that r < n, then P(n, r) represents the number of all possible arrangements or permutations of n distinct objects taken r at a time.

It can be represented as nPr = \(\frac{n!}{(n-r)!}\).

The combination is defined as “An arrangement of objects where the order in which the objects are selected does not matter.” 

nCr = \(\frac{n!}{r!(n-r)!}\) ,  when n < r

Where n = distinct object to choose from

C = Combination

r = spaces to fill

Calculation:

Vowels = 2

Consonants = 5

Total Alphabets = 7

Since 4 letter words must include 2 vowels, we don't need to select them, and the rest of the 2 letters will be taken from 5 consonants.

Number of ways of selecting 2 letters from 5 consonants = 5C2 = 10

Arrangement of all 4 letters will be given by 4! = 24 ways

Total number of arrangements = 5C2 × 4! = 10 × 24 = 240 ways

∴ The total number of words that can be formed is 240.

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Exercise :: Permutation and Combination - General Questions

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13.

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

A. 10080 B. 4989600 C. 120960 D. None of these Answer: Option C Explanation: In the word 'MATHEMATICS', we treat the vowels AEAI as one letter. Thus, we have MTHMTCS (AEAI). Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different. Number of ways of arranging these letters = 8! = 10080. (2!)(2!) Now, AEAI has 4 letters in which A occurs 2 times and the rest are different. Number of ways of arranging these letters = 4! = 12. 2! Required number of words = (10080 x 12) = 120960.

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Exercise :: Permutation and Combination - General Questions

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7.

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

Answer: Option D Explanation: Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it. The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place. The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it. Required number of numbers = (1 x 5 x 4) = 20.

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