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# earlier probability question by hung ho probability question

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Current Student
Joined: 07 Aug 2009
Posts: 49
earlier probability question by hung ho probability question [#permalink]

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03 Sep 2009, 12:07
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0% (00:00) correct 100% (02:29) wrong based on 1 sessions

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Quote:
There is a word AUSTRALIA. Four letters are taken at random. How many different words can be formed that have two vowels and two consonants?

Solution:
Vowels: A, U, A, I, A. Distinguish vowles: A, U, I.
Consonants: S, T, R, L

Outcomes with two vowels and two consonants:
1) The two vowels are different: C(3,2)*C(4,2)
2) The two vowels are the same: C(1,1)*C(4,2)
Total outcomes = 18+6=24

I dont understand the 2nd part for two vowels that are the same:2) The two vowels are the same: C(1,1)*C(4,2)
Manager
Joined: 18 Jul 2009
Posts: 169
Location: India
Schools: South Asian B-schools
Re: earlier probability question by hung ho probability question [#permalink]

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04 Sep 2009, 00:06
we have
5 Vowels - 3A / U / I
4 Consonants - S / T /R / L

Hence we will select two vowels out of 5 and two consonants out of 4...

we can do this in [5C2/3!]*4C2 but thus formed 4 letter word can be arranged in 4! ways

Thus => [5C2/3!]*4C2*4! = 4!*10 = 240 words

if u like my post consider it for Kudos!
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Bhushan S.
If you like my post....Consider it for Kudos

Manager
Joined: 19 Jul 2009
Posts: 53
Schools: McIntire, Fuqua, Villanova, Vandy, WashingtonU
Re: earlier probability question by hung ho probability question [#permalink]

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06 Sep 2009, 17:41
bhushan252 wrote:
we have
5 Vowels - 3A / U / I
4 Consonants - S / T /R / L

Hence we will select two vowels out of 5 and two consonants out of 4...

we can do this in [5C2/3!]*4C2 but thus formed 4 letter word can be arranged in 4! ways

Thus => [5C2/3!]*4C2*4! = 4!*10 = 240 words

if u like my post consider it for Kudos!

1. [5C2/3!] is not even an integer
2. You can multiply by 4! ONLY in case when two vowels are different. AABC does not not create 24 words, only 12:
AABC
ABAC
ABCA

AACB
ACAB
ACBA

BAAC
BACA
BCAA

CAAB
CABA
CBAA

3. IMO answer is 3C2 * 4C2*4! + 4C2 * 4! / 2!
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...
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Manager
Joined: 11 Sep 2009
Posts: 129
Re: earlier probability question by hung ho probability question [#permalink]

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18 Sep 2009, 00:55
AUSTRALIA

Consonants (4): S, T, R, L
Vowels (5): A, A, A, I, U

This problem needs to be seperated into two cases:

1) The two vowels are distinct

4C2 ways to choose the consonants
3C2 ways to choose the vowels
4! ways to arrange these 4 letters

Total # of ways: 4C2 * 3C2 * 4! = 432

2) Using A twice for both vowels

4C2 ways to choose the consonants
1C1 ways to choose the vowels (its A, A)
4!/2! ways to arrange these letters

Total # of ways: 4C2 * 1C1 * 4!/2! = 72

Total: 432 + 72 = 504
Re: earlier probability question by hung ho probability question   [#permalink] 18 Sep 2009, 00:55
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