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How many 5-digit numbers can be formed from the digits

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How many 5-digit numbers can be formed from the digits [#permalink]

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11 Nov 2007, 03:53
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How many 5-digit numbers can be formed from the digits 2,3,5,6,8,9 if no digit can be used more than once in a number ? How many even numbers can be formed ?

In how many ways can 4 consonants and 3 vowels be arranged in a row (a) so that the 3 vowels are always together, (b) so that the first and the last places are occupied by consonants.
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11 Nov 2007, 05:04
For question number 1, answers are 360 and 940 respectively

For question number 2, answers are 720 and 1440 respectively

Correct me if I am wrong, and let me know if I am right; I will provide the solution both conventional and shortcut

Amardeep
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11 Nov 2007, 08:08
I think the answer fo rthe first question is 720.

You have 6 digits and 5 spots. So, the first digit can be any of the 6, the second digit can be any of the remaining 5, and so on:

6 x 5x 4 x 3 x 2 = 720
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11 Nov 2007, 09:16
How many 5-digit numbers can be formed from the digits 2,3,5,6,8,9 if no digit can be used more than once in a number ?

How many even numbers can be formed ?

or simply because there are equal even and odd numbers out of 720 half will be even and half odd = 360

In how many ways can 4 consonants and 3 vowels be arranged in a row (a) so that the 3 vowels are always together?

Answer = consider the three vowels as one letter
so we get 5*4*3*2*1 = 120

(b) so that the first and the last places are occupied by consonants.

here condition "a" does not continue

I hope i m right
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Re: more perm to solve [#permalink]

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14 Nov 2007, 23:46
Ravshonbek wrote:
In how many ways can 4 consonants and 3 vowels be arranged in a row (a) so that the 3 vowels are always together, (b) so that the first and the last places are occupied by consonants.

(a) consider 3 vowels as one package, 4 consonants as 4 packages. Arranging these 5 packages, we have 5! ways. Besides, 3 vowels arrange within one package in 3! ways. So in total, we have 3!*5! ways to arrange as requested.

(b) there're C(4,1) alternatives for first place; then, there're C(3,1) alternatives for last place. We're left with 2 consonants and 3 vowels to place in between. There're 5! ways to arrange these 5 entities => The number of way for us to arrange as requested is C(4,1)*C(3,1) * 5!
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Re: more perm to solve [#permalink]

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15 Nov 2007, 06:40
laxieqv wrote:
Ravshonbek wrote:
In how many ways can 4 consonants and 3 vowels be arranged in a row (a) so that the 3 vowels are always together, (b) so that the first and the last places are occupied by consonants.

(a) consider 3 vowels as one package, 4 consonants as 4 packages. Arranging these 5 packages, we have 5! ways. Besides, 3 vowels arrange within one package in 3! ways. So in total, we have 3!*5! ways to arrange as requested.

(b) there're C(4,1) alternatives for first place; then, there're C(3,1) alternatives for last place. We're left with 2 consonants and 3 vowels to place in between. There're 5! ways to arrange these 5 entities => The number of way for us to arrange as requested is C(4,1)*C(3,1) * 5!

Very nice. Thanks.
Re: more perm to solve   [#permalink] 15 Nov 2007, 06:40
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