From 7 consonants and 5 vowels, how many words can be formed

Question

From 7 consonants and 5 vowels, how many words can be formed consisting of 4 different consonants and 3 different vowels? The words need not have meaning.

OA:
 1,764,000

Bunuel · Accepted Answer

The # of ways to choose 4 different consonants out of 7 is  C^4_7=35 ;
The # of ways to choose 3 different vowels out of 5 is  C^3_5=10 .

Thus the # of ways to choose 4 different consonants and 3 different vowels is  35*10=350 .

Finally these 7 letters can be arranged in 7! ways, therefore the total # of different words is  350*7!=1,764,000 .

Hope this helps.

Yurik79 · Answer

Number of ways to choose 4 cons out of 7*Number of ways to choose 3 vowels out of 5---->4C7*3C5=(7*6*5*4/1*2*3*4)*(5*4*3/1*2*3)=350

krisrini · Answer

Since the order of the letters are important we get
7P4*5P3 = 4200

Itzrevs · Answer

sorry guys both of you didn't get the OA.. Lets give it another try ;-)

upamanyu · Answer

4 consonants can be chosen in 7*6*5*4 ways

3 vowels can be chosen in 5*4*3 ways

Total no. of alternatives =7*6*5*4*5*4*3=50,400

crazyfin · Answer

u have put the right formula but calculated it incorrectly. ur answer shud also be 50400....

Itzrevs · Answer

Yurik79, I didn't the same way as you did. But thats not the answer.

I shall wait for few more people to jump in and post their answers & explanations..

Will post the OA soon.

krisrini · Answer

Thanks crazyfin. The compuatation should have worked to 50400, I am not sure how I made this mistake, thanks much for pointing out.

Itzrevs · Answer

Guys, The OA is  1,764,000 

OE :

The 4 different consonants can be selected in 7C4 ways
The different vowels can be selected in 5C3 ways
The resulting 7 different letters ( 4 consonants & 3 vowels) can be arranged themselves in 7P7 = 7! ways..

So the answer is 7C4 * 5C3 * 7! = 35 *10 * 5040 = 1764000

Yurik79 · Answer

It seemed so simple at first look)))Good one!10x

Lobro · Answer

Because each word has to consist of  different  consonants or vowels, the correct answer is (7*6*5*4)*(5*4*3) and not 7^4*5^3. I used bars to distinguish the numbers representing consonants and vowels to make it more clear.

Aldossari · Answer

I do not think you will see like this question in the test, unless the answer choices contain 350 * 7!.

but this is a good question to practise both concept - combination and permutation- in one question.

shashankprasad · Answer

Hi, I still don't get whats wrong with 7P4*5P3?

Bunuel · Answer

That's because we are arranging all 7 letters together.

For example, consider simpler case: how many 4-letter words you can make from 2 consonants {b, c} and two vowels {a, e}. If you arrange them first ({b, c} and {c, b} for consonants and {a, e} and {e, a} for vowels) you'll get 2*2=4 words, which would be wrong. The correct answer is 4!=24.

shaderon · Answer

Hello there

Are there ways to come up with sub-types of this particular question?

For example,

A case with 3 distinct vowels & 4 distinct consonants.
A case where the ORDER of the letters DOES NOT matter (OR) a case where ORDER matters.
etc.

OR is there a particular link where such questions are discussed?

Thanks in advance.

Bunuel · Answer

The question at hand is where we have distinct letters and the order does matters. If the order does not matter the answer would be C^4_7*C^3_5=35*10=350 . Theory on Combinations: math-combinatorics-87345.html DS questions on Combinations: search.php?search_id=tag&tag_id=31 PS questions on Combinations: search.php?search_id=tag&tag_id=52 Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html Hope this helps.

282552 · Answer

Hi Bunuel
What is wrong with my approach.
Can you please correct me : 
Imagine 7 boxes now let's say we are filling first 4 boxes with consonants .First box can be filled with 7 ways.Second with 6 and so on.
Similarly for 3 vowel boxes : first box can be formed with 5 ways , second with 4 ways and third with 3 ways :

So total number of ways =7*6*5*4*5*4*3
And now these letters can be arranged in 7! ways to have different combination.
I know I am wrong.
Can you please correct me.

Thanks in Advance.

Bunuel · Answer

Short answer would be that 7*6*5*4 as well as 5*4*3 will contain repetitions. The number of way to choose 4 letters out of 7 is NOT 7*6*5*4 it's  C^4_7=\frac{7!}{4!3!}=35  and the number of way to choose 3 letters out of 5 is NOT 5*4*3 it's  C^3_5=\frac{5!}{3!2!}=10 .

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From 7 consonants and 5 vowels, how many words can be formed

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From 7 consonants and 5 vowels, how many words can be formed

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