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

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.

ⁿP_r= (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.

ⁿC_r = 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