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Combinatorics: If a code word is defined to be a sequence of

by II » Sat May 10, 2008 12:41 am
If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 5-letter code words to the number of 4-letter code words?

A) 5 to 4
B) 3 to 2
C) 2 to 1
D) 5 to 1
E) 6 to 1

Interested in finding out the various approaches to solve this one.
Also if you had to make an educated guess on this ... how would you go about it ?

Thanks in advance.
II
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by codesnooker » Sat May 10, 2008 4:45 am
Answer : 6 to 1

Simple and Straight approach, no need to have prior knowledge of combination or permutation.

Let's check the count of number of ways to chose 4 letters.

Its already given in the question that we need form the group with different letters.

So for place 1, how many letters do you have? 10
So for place 2, how many letters do you have? 9 (one letter is already out)
So for place 3, how many letters do you have? 8 (two letter are already out)
So for place 4, how many letters do you have? 7 (three letter are already out)

Totals ways for group of four letters = 10 X 9 X 8 X 7

Similarly, total ways for forming the group of 5 letters = 10 X 9 X 8 X 7 X 6

That is,
Group (5):Group (4) :: 10 X 9 X 8 X 7 X 6 : 10 X 9 X 8 X 7

Group (5):Group (4) :: 6 : 1

So without any special knowledge you can solve it in seconds and no need to make any educated guess in such kind of questions.
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by ohwell » Mon Nov 10, 2008 8:04 pm
codesnooker is right on, but if anyone is interested in a little bit of calculations...

The question is about permutations. In how many ways can we create a 4 character code and a 5 character code.

The formula for permutation (P) is:
P(n,k)= n!/(n-k)! (and if anyone is interested in any relationship with combinations which is not what is being asked here though: P(n,k) = r! * nCr)

Using the above yields the same as codesnooker because a 5 character code can be chosen 10! (10-5)! ways or 10*9*8*7*6

and a 4 character code can be chosen 10! / (10-4)! ways = 10*9*8*7

dividing the two yields 6/1.
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Re: Combinatorics: If a code word is defined to be a sequenc

by logitech » Mon Nov 10, 2008 8:14 pm
II wrote:If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 5-letter code words to the number of 4-letter code words?

A) 5 to 4
B) 3 to 2
C) 2 to 1
D) 5 to 1
E) 6 to 1

Interested in finding out the various approaches to solve this one.
Also if you had to make an educated guess on this ... how would you go about it ?

Thanks in advance.
II
You know how I always start explaining something: KISS

Keep it simple stupid!

Lost of people approach problems in this forums as they are sitting in a 4 hrs Calculus test in a college. This is BUSINESS SCHOOL test, and one needs to think like a businessman.

What will be the difference between COMB 5 and COM4 ?

you will use all the combinations in COM4 inside the COM5 right ? But you will also have 1 more number to play with that is left out by COM4.

How many numbers are left out in COM4 ? 6

This is your guess my friend.

10 - 4 = 6 is the solution. Hence, E
LGTCH
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by cramya » Mon Nov 10, 2008 8:16 pm
OOhh. We got a fancy swancy business man here in this forum people :-)

I feel privilaged to be part of this forum now!
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by logitech » Mon Nov 10, 2008 8:21 pm
cramya wrote:OOhh. We got a fancy swancy business man here in this forum people :-)

I feel privilaged to be part of this forum now!
Keep it simple stupid!
LGTCH
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by cramya » Mon Nov 10, 2008 8:23 pm
Keep it simple stupid!
Right back at u

U too "Keep it simple stupid" :-)
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by ohwell » Mon Nov 10, 2008 9:31 pm
I am all for KISS, now worries there mate...I think in this case we are all getting a bit too excited, as it takes far more time to respond to these questions from others and to respond again like this. While I am not entirely understanding the way you approached it, my approach really did not take me very much time, maybe 20 seconds. I would think this is well within the average per question...
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by cramya » Mon Nov 10, 2008 9:37 pm
Well put Ohwell!

All minds are not alike.. All we need to worry about is if we can complete a problem in 2 minutes irrespective of the approach taken(easier said than done for all of us). I am sure there will be easier problems that can be done in less than 2 minutes if u have the opener (starting approach for a problem) and tougher ones that may take more than 2 minutes even with a opener.

I agree calculating using permutation forumla can be done quick also since most of the terms cancel wiht the excpetion of 6 in NR adnd 1 in DR
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Re: Combinatorics: If a code word is defined to be a sequenc

by fighting_cax » Mon May 11, 2009 4:12 am
logitech wrote:
II wrote:If a code word is defined to be a sequence of different letters chosen from the 10 letters A, B, C, D, E, F, G, H, I, and J, what is the ratio of the number of 5-letter code words to the number of 4-letter code words?

A) 5 to 4
B) 3 to 2
C) 2 to 1
D) 5 to 1
E) 6 to 1

Interested in finding out the various approaches to solve this one.
Also if you had to make an educated guess on this ... how would you go about it ?

Thanks in advance.
II
You know how I always start explaining something: KISS

Keep it simple stupid!

Lost of people approach problems in this forums as they are sitting in a 4 hrs Calculus test in a college. This is BUSINESS SCHOOL test, and one needs to think like a businessman.

What will be the difference between COMB 5 and COM4 ?

you will use all the combinations in COM4 inside the COM5 right ? But you will also have 1 more number to play with that is left out by COM4.

How many numbers are left out in COM4 ? 6

This is your guess my friend.

10 - 4 = 6 is the solution. Hence, E
I'm a bit confused with this approach...

How were you able to get the values of "COMB 5" and "COM4"?
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by su_gmat » Wed Feb 15, 2012 7:42 am
It is very simple..

ration=no way to pick 5 different letters /way to pick 4 different letters

10p5/10p4=1:6

thanks,
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