Aptitude - Permutation and Combination - Discussion

This question can be solved by two methods
1) using combinations:

we have to select 3 consonants from 7 and 2 vowels from 4,
we will use 7C3 x 4C2, this implies that for eg. consonants are B C D F G H J, then
if BCD is there then CBD and DBC won't be there and so on..
but they will make different words, so
we arrange them later on by multiplying by 5! ..
eg. one combination will be BCD AE this can be arranged in 5! ways and so on...
so ans. is 7C3 x 4C2 x 5! = 25200

2) using permutations:

unlike previous method if we use 7P3 we also consider BCD, CBD and all possible arrangements...

then we get combinations of the type CCCVV (C- consonant, v- vowel)
but they can be arranged in 5! ways eg- CVCVC, VVCCC, etc..
but using permutation, we have already accounted for those combinations in which all 3 consonants occur together and 2 vowels occur together.. so to eliminate those cases, we need to divide it by the repetitions i.e. divide by 3!2! (for consonants and 2 for vowels) and then multiply this with the initial result..
so ans is 7P3 x 4P2 x 5!/(2!3!) = 25200

tried to explain as clearly as possible..hope i was of some help..

@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.

Simplify problem.

3 consonants BCD.
3 vowels AEI.

Choose 2 consonants, no repetition.
Choose 2 vowels, no repetition.

Form different words from the chosen consonants and vowels.

Choose 2 consonants, 3 ways - BC, BD, CD.
Choose 2 vowels, 3 ways - AE, AI, EI.

Different groups with 2 consonants and 2 vowels, 3x3 = 9.

BCAE, BCAI, BCEI, BDAI, BDAI, BDEI, CDAE, CDAI, CDEI.

Different words from 4 character set such as BCAE, 4! = 4x3x2x1 = 24.

BCAE, BCEA, BACE, BAEC, BECA, BEAC.
CAEB, CABE, CEAB, CEBA, CBAE, CBEA.
ABCE, ABEC, AEBC, AECB, ACBE, ACEB.
EBCA, EBAC, ECBA, ECAB, EABC, EACB.

Total 4 character words from 2 different consonants and 2 different vowels would be 9x24 = 216.

Apply to above.

7C3 x 4C2 x 5! = 25200.

More difficult would be when repetitions are allowed in consonants and vowels.

Explanation is actually simple. Suppose if they asked to find the number of words possible using five different letters and you had no constraints of consonants and vowels.

Then the answer would obviously 5!. So, we know you can form 5! words from five letters.

Now, in the problem, there are not just 5 letters, but much more - 7 consonants and 4 vowels. The 3 consonants can be selected in 7C3 ways. For each of these, we can select 2 vowels in 4C2 ways.

So, the 5 letters themselves can be selected in 7C3X4C2 ways.

Note that each of these 7C3X4C2 are a different set of five letters. A given set of 5 letters can be arranged in 5! ways.

So, 7C3X4C2 sets of 5 letters can be arranged in 7C3X4C2X5! ways.

Hence the answer is 25,200

@ Anil: The basic idea here is that first you pick out 3 consonants out of 7 and 2 vowels out of 4. Then you take the 5 letters that you have and make a word out of it. To pick out 3 consonants out of 7 you use the combination without repetition formula which states n!/(n-r)!r! which gives you 7C3 and the same for vowels gives you 4C2.

Once you have done that, you still need to make a word of it. You have 5 letters now that can be arranged into 5! ways to make a word so you multiply the three values since each step is dependent on the step before to make the final word. The answer comes out to be 35 x 6 x 120 respectively for each step mentioned chronologically.

It's simple,

Select 3 consonants out of B,C,D,F,G,H,J (assume these 7 consonants) in 7P3 ways.
So, [ B C D ] is one selection out of 7P3 ways right?
but, [ C B D] , [ C D B ] , [ B D C ] , [ D C B ] , [ D B C ] , all these contain same consonants.

This is why you use the combination, to form groups of 3 consonants out of 7 consonants.
Same goes for vowels : 4P2 ways to get groups of 2 vowels out of 4 vowels.
total groups of consonants and vowels : 7P3 * 4P2.
Now, we need to form words of 5 letters consisting of vowels and consonants.
How many ways? In 5! ways.

Thus, 7P3 * 4P2 * 5! = 25200 ways.

This cannot be a permutation problem because the first thing we ought to do is select 4 consonants out of 7 and 2 vowels out of 4.

Let us approach the question by assigning actual alphabets, and reducing the quantity. Suppose there are 3 consonants B, C and D and 2 vowels A and E, and we have to form a word using 2 consonants and 1 vowel.

The required words are: ABC, BAC, BCA, ACB, CAB, CBA, ACD, CAD, CDA, DCA, DAC, ADC.....the sum would be 36.

Let us now do the sum using formula:
3P2*2P1=12, which is clearly not the answer.

210 is correct. What is asked in the problem is how many words that have 3 consonants and 2 vowels can be formed from the given 7 and 4.

Basically all words that contains 3 consonants no matter how you arranged the 3 consonants that you picked is considered as one, same with the vowels. Meaning QEWAR is same with WEQAR, hence counted as 1 sample.

Unless you choose another consonant say QEWAD or another vowel QIWAR then it will be a different sample.

Dear friends

Keep in mind first that it is a permutation problem becoz the order of letters matter in order to make different words. Secondly out of 7 consonants we have to select 3 so when we select a consonant out of 7 it will not again be considered to be selected so the repetition is not allowed. Now the formula for Permutation without repetition is (n!)/(n-r)!. So it will be 7P3*4P2.

=(7!/4!)*(4!/2!)=2520 (not 25200).

Hi,

I do agree with Sachin Jain as it is a permutation query. We can arrange it in any way using using 3 consonants out of 7 & 2 vowels out of 4.

Ex: If there is bcdfghj then we can do it bcd n also cbd so it is permutation type not combination so the answer should be 7P3 * 4P2 = 2520

But again we have to arrange these 2 groups together. So I'm not getting form this point.

Can anyone clear the same?