Aptitude - Permutation and Combination - Discussion

If the question is how many words we can form then why need to find the number of ways?

This question mentions it 7C3=> 7x6x5x4/6.
7x5x4.
= > 35x4,
= > 140,
= > 140 x 6x 120,
= > 140x 720 ways.
= > 100800 ways.

@All.

Here, it should be done with permutation at the beginning instead of using combinatation, which just make no sense. The calculation starts from the permutation of 7!/4! * 4!/2!. Once this is finished, we still have to figure out a way to arrange these 5 letters. We can start from the simplest pattern, which is 3 consonants next to each other on the right side and 2 vowels next to each other on the left side. Since we have already worked out the all the possible changes for every single letter in the previous step, we can know start to work on how many different patterns these five letters can be possbily arranged. There is 4 arrangements if the the first letter is vowel and 3 arrangements if the second letter is vowel. (the one that is already observed does not count).

Eventually, you will get 1+2+3+4=10 different sets of letters. You should then mutiply the result from the perious permutation caculation (2520) with 10, because each set of letters still has 2520 different arrangements.

The result will finally be:

7!/4! * 4!/2! * (1+2+3+4）=25200，.

I hope, this will calrify the confusion.

Here we can directly permute 7p3*4p2=2520 so this gives us 3 vowels say "aei" and 2 consonants say bc so aeibc we can then calculate the number of groups made by the 2 consonants or 3 values from this group of 5 letters chosen in total so the solution is 7p3*4p2*5c2 = 7*6*5*4*3*(5*4)/2=2520 *10 = 25200.

Why at 7p3 4 is excluded can anyone help me with this?

Why we want to multiply with 5! Please tell me.

Hi, @All.

I am just confused with committee problems and this question in committee problems we are not doing this 5! thing if a committee of 5 people is to be made, we are just selecting the men and women out of given men and women and multiplying why it is not similar in this case.

Can anyone help to understand the problem? Please help me.

Why do 5!?

What if the consonants or vowels have repetition? Does it necessary to multiply with 5!?

This is a question of permutation, as the order in which letters are arranged changes the final result.

Therefore = 7P3*4P2 = 2520.

However, there are 10 different ways that the consonants and vowels can be arranged.

So, the answer is 2520*10,
=2520.

7P3 * 5P2 =2520.

Then considering combining them accounting for repetitions we get = 5!/(3!*2!)=10.
Therefore =2520*10=25200.