cata1yst wrote:ILULA08 wrote:After removing 5 vowels from 26 Alphabets we have 21 consonants
i) Selecting 4 vowels out of 5= 5C4
ii)Selecting 13 consonants our of 21= 21C13
These 17 selected alphabets can arrange themselves in 17! ways.
So total number of ways- 5C4*21C13*17!
What is derived if you don't multiply the extra 17! ???
I solved everything else, but left off 17! since I already thought I got the answer.
Let's look at an analagous question to understand the difference:
1) Bob is having a party and is inviting a total of 17 friends. Bob will invite 4 men and 13 women to his party. If Bob has 5 male friends and 21 female friends, how many different groups of friends could he invite to his party?
In this question, we're selecting 4 out of 5 men and 13 out of 21 women, so the answer would be:
5C4 * 21C13
(remember, when we make MULTIPLE selections we MULTIPLY the results).
Next, let's add an extra dimension to the question:
2) Bob is hosting a party with 17 friends. At the party, all 17 guests will dance in a conga line (
https://www.youtube.com/watch?v=zI5mfE9dhFU). In how many different ways can the guests be arranged for the dance?
In this question, we're arranging 17 people, so the answer would be:
17P17 = 17!
Now let's combine the two questions:
3. Bob is having a party and is inviting a total of 17 friends. Out of Bob's 5 male friends and 21 female friends, he will invite 4 men and 13 women to his party.
At Bob's party, all 17 of his friends will dance in a conga line. Out of Bob's total of 26 friends, how many different conga lines can be formed?
In this question, we're not only selecting 4 out of 5 and 13 out of 21, we're also arranging those that we select. So, our final answer is:
(# of selections) * (# of arrangements)
= (5C4*21C13) * (17!)
