Aptitude - Permutation and Combination - Discussion

Discussion Forum : Permutation and Combination - General Questions (Q.No. 12)
12.
How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
40
400
5040
2520
Answer: Option
Explanation:

'LOGARITHMS' contains 10 different letters.

Required number of words = Number of arrangements of 10 letters, taking 4 at a time.
= 10P4
= (10 x 9 x 8 x 7)
= 5040.

Discussion:
68 comments Page 4 of 7.

Shareef said:   10 years ago
Why 4 is taken?

Sanjana said:   10 years ago
Can the clear explanation of this problem can be given by someone? Since I dint get it.

Haribabu said:   1 decade ago
Why are you taking 10c4 so many ways like 10c2, 10c5?

If I take total 10 letters, so I calculate 10! = (10*9*8*7*6*5*4*3*2*1).

So please explain clearly.

Onkar said:   1 decade ago
We can't use permutation here because they have mentioned that no repetition is permissible and in 4 number word only possible solution is 210. 10C4 = 210.

Ankur said:   1 decade ago
It should be done in simple way i.e. 10 c4 = 210 ways then 4 letters can be written in 4! i.e 24 therefore 210*24 = 5040.

Nandu said:   1 decade ago
@Santosh.

If you see "SELECT" then use "ncr" and if you see "ARRANGE" use npr.

No confusion.

Santhosh said:   1 decade ago
I have so much confusion at the answering problem,
"where we use the permutation formula and combination formula"?

Please anyone tell me.

Ashutosh Sharma said:   1 decade ago
Which formula is used to solve this problem?

Pradeep said:   1 decade ago
Its very easy, no need to get confused with permutation.

Divide the statement in two part.

From given 10 character, we have to choose 4 of them only.
So, 10C4 = 10*9*8*7/4! = 210.

Now we have to arrange there 4 character.
So, it would be simply 4! because there is no repetition of word.

Total - 210*24 = 5040.

Moral - choose 4 word out of 10 then arrange them by 4!.
(1)

Motunrayo said:   1 decade ago
Please I want to know when to use combination and permutation.


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