How many words of 5 letters each can be formed each containing 3 consonants and 2 vowels?

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How many words can be formed, each of $2$ vowels and $3$ consonants from letters of the word "DAUGHTER"

What my textbook has done: it has first taken combinations of vowels and then consonants then multiplied them altogether. Now for each combination of words they can be shuffled in $5!$ ways, so multipliying by $5!$ we get the required answer.

My question is: why has the book used combinations instead of permutations while selecting vowels and consonants?

Thanks.

JKnecht

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asked Jun 4, 2015 at 19:00

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All the letters are different, so that makes things easier.

Pick the two vowels ($_3C_2$) and pick the three consonants ($_5C_3$) and then pick what order they go in $(5!)$. So the answer is $3 \cdot 10 \cdot 120 = 3600.$

You take combinations of the vowels and consonants because the order of them doesn't matter at that point. You order them in the last step, after you've chosen which ones go in your five-letter word.

In other words, it doesn't matter that I pick $A$, then $U$, instead of $U$, then $A$. It just matters that I picked the set $(A,U)$.

answered Jun 4, 2015 at 19:12

JohnJohn

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Your query why not permutation first ? As, you have to make words of length=$5$. And of these $5$, $2$ are vowels and $3$ consonants. Since, you have to first get those $2$ vowels and $3$ consonants to make the desired word. So first operation has to be combination(selection operation), which will select $2$ vowels out of $3$ vowels(A,E,U) and then you have to select 3 consonants out of $5$(D,G,H,T,R). And they need to be multiplied, as there can be many such combinations i.e $C(3,2)*C(5,3)$. Now that you have formed $5$ letter word. These letters can be arranged among themselves to make different words. Hence, you need to apply permutation(arrangement) i.e. $5!$, making final result= $C(3,2)*C(5,3)*5!$.

answered Jun 4, 2015 at 19:40

user2016963user2016963

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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.

How many words of 3 vowels and 6 consonants can be formed taken from 5 vowels and 10 consonants?

Answer:

Total no. of vowels = 5 

Total no. of consonants = 10

No. of words with 3 vowels and 6 consonants

3 vowels can be selected from 5 vowels = 5C3 ways = n!/(n-r)!r!= 5!/(5-3)!3! =10 ways

6 consonants can be selected from 10 consonants = 10C6ways =  n!/(n-r)!r! = 10!/(10-6)!6! = 210 ways

Total selection = 5C3  ×  10C6

Now, 9 letters in each selection can be arranged in 9! ways

Total no. of words = 5C3 ×  10C6  × 9!

                                      = 10  ×  210 × 9!

                            = 2100 × 9!

                            = 762,048,000 words

Similar Questions

Question 1: If 5 vowels and 6 consonants are given, then how many 6 letter words can be formed with 3 vowels and 3 consonants?

Answer:

Total no. of vowels = 5

Total no. of consonants = 6

The no. of  6 letter words with 3 vowels and 3 consonants

3 vowels can be selected from 5 vowels = 5C3 ways = n!/(n-r)!r!= 5!/(5-3)!3! =10 ways

3 consonants can be selected from 6 consonants = 6C3 ways =  n!/(n-r)!r! = 6!/(6-3)!3! = 20 ways

Total selection = 5C3  × 6C3

Now, 6 letters in each selection can be arranged in 6! ways

Total no. of 6 letter words = 5C3 × 6C3  × 6!

                                         = 10 × 20 × 6!

                                         = 200 × 6! 

                                         = 1,44,000 words   

Question 2: How many different words each containing 3 vowels and 5 consonants can be formed with 5 vowels and 19 consonants?

Answer:

Total no. of vowels = 5

Total no. of consonants = 19

No. of words with 3 vowels and 5 consonants

3 vowels can be selected from 5 vowels = 5C3 ways = n!/(n-r)!r!= 5!/(5-3)!3! =10 ways

5 consonants can be selected from 19 consonants = 19C5 ways =  n!/(n-r)!r! = 19!/(19-5)!5! = 11,628 ways

Total selection = 5C3  × 19C5

Now, 8 letters in each selection can be arranged in 8! ways

Total no. of  words = 5C3 × 19C5 × 8!

                             = 10 × 11,628 × 8!

                             = 116280 × 8!

                             = 4,688,409,600 words   

Question 3: How many different words each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants?

Answer:

Total no. of vowels = 5

Total no. of consonants = 17

No. of different words with 2 vowels and 3 consonants

2 vowels can be selected from 5 vowels = 5C2 ways = n!/(n-r)!r!= 5!/(5-2)!2! =10 ways

3 consonants can be selected from 17 consonants = 17C3 ways =  n!/(n-r)!r! = 17!/(17-3)!3! = 680 ways

Total selection = 5C2 × 17C3

Now, 5 letters in each selection can be arranged in 5! ways

Total no. of  words = 5C2 × 17C3 × 5!

                              = 10 × 680 × 5!

                              = 6800 × 5!

                              = 8,16,000 words     

Question 4: How many different words each containing 2 vowels and 3 consonants can be formed with 4 vowels and 7 consonants?

Answer:

Total no. of vowels = 4

Total no. of consonants = 7

No. of different words with 2 vowels and 3 consonants

2 vowels can be selected from 4 vowels = 4C2 ways = n!/(n-r)!r!= 4!/(4-2)!2! = 6 ways

3 consonants can be selected from 7 consonants = 7C3 ways =  n!/(n-r)!r! = 7!/(7-3)!3! = 35 ways

Total selection = 4C2 × 7C3

Now, 5 letters in each selection can be arranged in 5! ways

Total no. of  words = 4C2 × 7C3 × 5!

                             = 6 × 35 × 5!

                             = 210 × 5!

                             = 25,200 words    

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