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# Rules 1. Time yourself 2. Work your solution on a seperate

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CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 702 [0], given: 781

Rules 1. Time yourself 2. Work your solution on a seperate [#permalink]  09 Mar 2004, 15:12
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Rules

1. Time yourself
2. Work your solution on a seperate sheet of paper

I say it again, we request your timings for different problems. this will help us in our project. you can just say >2 min or <2 min...something like that.

How many four letter distinct initials can be formed using the alphabets of English language such that the last of the four words is always a consonant?

[ Consonant => all aphabets except the vowels A,E,I,O and U ]

1) (26^3)*(21)
2) 26*25*24*21
3) 25*24*23*21
4) 25^3*21
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

less than 10 seconds

26^3 * 21

I dont understand what you mean by distinct initials.

is AAAA a distcint initial or not ?

Last edited by anandnk on 09 Mar 2004, 16:44, edited 1 time in total.
Senior Manager
Joined: 05 May 2003
Posts: 427
Location: Aus
Followers: 2

Kudos [?]: 5 [0], given: 0

3) 25*24*23*21
I assumed distinct initial means ABCD without repeating them in the same set.

I took 1 min.
Senior Manager
Joined: 06 Dec 2003
Posts: 366
Location: India
Followers: 1

Kudos [?]: 4 [0], given: 0

Another vote for 25*24*23*21;

Explanation:
First select consonant for 4th place, one can do that in 21 ways. now not to repeat that consonant again select 1st, 2nd and 3rd place in 25, 24 and 23 ways respectively.

Cheers , Dharmin
CEO
Joined: 15 Aug 2003
Posts: 3469
Followers: 61

Kudos [?]: 702 [0], given: 781

Re: [#14] 2 Min. Challenge: Counting Methods - Language [#permalink]  10 Mar 2004, 00:04
praetorian123 wrote:
Rules

1. Time yourself
2. Work your solution on a seperate sheet of paper

I say it again, we request your timings for different problems. this will help us in our project. you can just say >2 min or <2 min...something like that.

How many four letter distinct initials can be formed using the alphabets of English language such that the last of the four words is always a consonant?

[ Consonant => all aphabets except the vowels A,E,I,O and U ]

1) (26^3)*(21)
2) 26*25*24*21
3) 25*24*23*21
4) 25^3*21

Edit: thanks kpadma, the correction has been done.

1) 26^3 * 21 is the answer. Good Discussion guys.
The question asks you to find out the number of distinct initials and not initials where the letters are distinct.
anandnk, is that clear now?
First letter can be arranged in 26 ways
Second letter can be arranged in 26 ways
Third letter can be arrranged in 26 ways

last one...NO vowels...so 26- 5 =21 ways

26^3 * 21 ways

thanks all
praetorian

Last edited by Praetorian on 10 Mar 2004, 13:33, edited 1 time in total.
Director
Joined: 03 Jul 2003
Posts: 655
Followers: 2

Kudos [?]: 22 [0], given: 0

Re: [#14] 2 Min. Challenge: Counting Methods - Language [#permalink]  10 Mar 2004, 07:52
praetorian123 wrote:
4) 25^3 * 21 is the answer. Good Discussion guys.

last one...NO vowels...so 26- 5 =21 ways

26^3 * 21 ways

thanks all
praetorian

Dear Praetorian,

Is it 25^3 * 21 or 26^3 * 21?
Senior Manager
Joined: 23 Sep 2003
Posts: 294
Location: US
Followers: 1

Kudos [?]: 2 [0], given: 0

Geethu and Dharmin,

Distinct initials = 26^3 * 21
Even if the first three letter are the same, the last consonant is sufficient to make all the initials distinct. BBBC is not the same as CBBB because we're talking initials.
Senior Manager
Joined: 05 May 2003
Posts: 427
Location: Aus
Followers: 2

Kudos [?]: 5 [0], given: 0

Thank you ndidi204.

anandnk, which probability book are you using. I haven't given any specific study to this topic separately. But you seem to give exact answers.
Senior Manager
Joined: 23 Sep 2003
Posts: 294
Location: US
Followers: 1

Kudos [?]: 2 [0], given: 0

Hey Geethu,

I recommend "Probability Without Tears" by Derek Rowntree. Great book. Completely explains probability/combinations/permuations in a very clear and straightforward manner. You can borrow it from your local library if you don't want to buy it.

It also gives several good examples.
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

Hi Geethu,

I didnt use any probability book. I just learnt it in this club by carefully analyzing every post. Very few times I visited
http://mathworld.wolfram.com/

Probability is one subject you will either hate or love.

Anand.
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