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by beat_gmat_09 » Wed Dec 08, 2010 9:50 pm
IMO C

different digits means - repetition is not allowed, create 5 digit numbers without repetition, the number should consist numbers 2,4 and 5 - there is no restriction on placing these numbers.
2,4,5 can be arranged in 3! ways. Remaining two digits can be arranged in 6*5 ways.
2,4,5 can occupy any position, number of ways to arrange - C(5,3) = 10 ways.
Total ways = 10*6*5*3*2 = 1800
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by Night reader » Wed Dec 08, 2010 10:32 pm
beat_gmat_09 wrote:IMO C

different digits means - repetition is not allowed, create 5 digit numbers without repetition, the number should consist numbers 2,4 and 5 - there is no restriction on placing these numbers.
2,4,5 can be arranged in 3! ways. Remaining two digits can be arranged in 6*5 ways.
2,4,5 can occupy any position, number of ways to arrange - C(5,3) = 10 ways.
Total ways = 10*6*5*3*2 = 1800
your answer is correct as 1800, but I am not for the solution way - please explain
C(5,3) = 10 ways
to my understanding the set is 5 and value is 3 i.e. n!/k!*(n-k) here must be n=3 and k=5 - you take an opposite
...
also 3! is combination of 2-4-5 only, we need to arrange this combination within the set of 5 places

...
as afterthought I solved this problem this way - plz, give your feedback

I start with constraints 2-4-5

they can be in any of the five, four and three places => 5 * 4 * 3
switch to not constraint => 6 digits left out of 9 digits, so 6 *
5 digits left out of 9 digits

also satisfies to different digit number 9*8*7*6*5 so 6*5 can be valid digits for the different digit numbers

5*4*3*6*5=1800
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by beat_gmat_09 » Wed Dec 08, 2010 11:22 pm
Night reader wrote: they can be in any of the five, four and three places => 5 * 4 * 3
I am not sure on this approach, perhaps Rahul or someone from the experts can shed light on its validity.
I'm describing my approach in more detail, hope you find this useful -

I) Fill the constraints first i.e. 5 digit numbers can comprise of 2,4 and 5. The remaining 2 numbers can be any which do not contain 2,4,5 i.e. Total 9 - 2,4,5 i.e. 3 = 6 numbers......... (1).
Now use simple permutation method -
1st can be filled in - 3 ways (you can start filling with 2,4,5 or can start with the remaining 6 numbers, as you wish, but keep in mind you also have to consider the possible arrangement of 2,4,5 in the 5 slots. 2,4,5 are not restricted to any place. Example - units, hundreds, thousands etc.)
Continuing filling from the restricted 3 numbers.
2nd can be filled in 2 ways, (rejecting 1 from the 1st place)
3rd can be filled in 1 way (reject 1 from 1st place and other from 2nd place)
4th number can be from the remaining 6 numbers (refer (1)), so total of 6 ways.
5th number can be filled in remaining 5 ways (reject 1 from 4th place)
So far total ways are - 3*2*1*6*5 = 180 ways.

II) As the position of numbers 2,4,5 is not restricted at any place we also have to consider their arrangements.
Sample numbers,which can be formed
- 245XX
- 452XX
.....
2XX45 can also be formed, or 2X4X5 can also be formed.
So there are more than 180 ways to arrange.
2,4,5 can be arranged in 5 slots in C(5,3) ways, or you can think this way - in how many ways can 3 numbers be put in 5 different slots - order does not matter here. Therefore C(5,3) ways = 10 ways.
Hence 10*180 = 1800 ways.
Perhaps your approach is correct but I cannot comment on it. I preferred what looked more logical to me.
HTH.
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by GMATGuruNY » Thu Dec 09, 2010 4:15 am
beat_gmat_09 wrote:IMO C

different digits means - repetition is not allowed, create 5 digit numbers without repetition, the number should consist numbers 2,4 and 5 - there is no restriction on placing these numbers.
2,4,5 can be arranged in 3! ways. Remaining two digits can be arranged in 6*5 ways.
2,4,5 can occupy any position, number of ways to arrange - C(5,3) = 10 ways.
Total ways = 10*6*5*3*2 = 1800
A 5-digit number offers 5 positions in which a digit can be placed. We should start with most restricted aspect of the problem: where to place the 2, 4 and 5.

Number of positions in which the digit 2 can be placed = 5.
Number of remaining positions in which the digit 4 can be placed = 4.
Number of remaining positions in which the digit 5 can be placed = 3.
5*4*3 = 60 ways to position the digits 2, 4 and 5.

Number of positions left = 2, number of digit choices left = 6 (since we can't reuse 2, 4 or 5).
Number of digit choices for the first remaining position = 6.
Number of digit choices for the last remaining position = 5.
6*5 = 30 ways to fill the last 2 positions.

Since to form the 5-digit integer, we need to combine our choices for the 2, 4 and 5 with our choices for the remaining 2 positions, we multiply the results above: 60*30 = 1800 possible integers.
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by diebeatsthegmat » Fri Dec 10, 2010 6:25 am
Night reader wrote:Five-digit numbers are composed of the digits 1, 2, 3, 4, 5, 6, 7, 8, 9. How many five-digit numbers with the different digits are composed with 2, 4, and 5 simultaneously?

A. 1100
B. 1600
C. 1800
D. 1900
E. 2000
the number could be abcde
there are five places for 2,4,5 so we have 5!/2!3!=10 combinations
2,4,5 has 3! =6 ways to replace its position in abcde
we will have 6 number left accept 2,4,5 to arrange the 2 positions left and those 2 positions must be with different number
so total =6*5 combination
total=6*5*10*6=1800
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by Night reader » Fri Dec 10, 2010 7:11 am
diebeatsthegmat wrote:
Night reader wrote:Five-digit numbers are composed of the digits 1, 2, 3, 4, 5, 6, 7, 8, 9. How many five-digit numbers with the different digits are composed with 2, 4, and 5 simultaneously?

A. 1100
B. 1600
C. 1800
D. 1900
E. 2000
the number could be abcde
there are five places for 2,4,5 so we have 5!/2!3!=10 combinations
2,4,5 has 3! =6 ways to replace its position in abcde
we will have 6 number left accept 2,4,5 to arrange the 2 positions left and those 2 positions must be with different number
so total =6*5 combination
total=6*5*10*6=1800
.....x245x
.....x425x
.....x542x
.....x524x
.....x452x
.....x254x

.....2x4x5
.....4x2x5
.....5x4x2
...

.....xx452
.....xx254
...

.....245xx
.....425xx
...

in total 24 combinations!
there are five places for 2,4,5 so we have 5!/2!3!=10 combinations
must be 3!/ 5!(3-5)! => factorial of -ve is undefined => use 3! multiplied by 2! combined set of combinations 3!*2!=24

so far not clear
2,4,5 has 3! =6 ways to replace its position in abcde
we will have 6 number left accept 2,4,5 to arrange the 2 positions left and those 2 positions must be with different number
so total =6*5 combination
total=6*5*10*6=1800
combining a combination of five numbers with the set of three numbers (calculated I don't understand why) with a combination of three numbers straight?
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by GMATGuruNY » Fri Dec 10, 2010 7:24 am
Night reader wrote:
diebeatsthegmat wrote:
Night reader wrote:Five-digit numbers are composed of the digits 1, 2, 3, 4, 5, 6, 7, 8, 9. How many five-digit numbers with the different digits are composed with 2, 4, and 5 simultaneously?

A. 1100
B. 1600
C. 1800
D. 1900
E. 2000
the number could be abcde
there are five places for 2,4,5 so we have 5!/2!3!=10 combinations
2,4,5 has 3! =6 ways to replace its position in abcde
we will have 6 number left accept 2,4,5 to arrange the 2 positions left and those 2 positions must be with different number
so total =6*5 combination
total=6*5*10*6=1800
.....x245x
.....x425x
.....x542x
.....x524x
.....x452x
.....x254x

.....2x4x5
.....4x2x5
.....5x4x2
...

.....xx452
.....xx254
...

.....245xx
.....425xx
...

in total 24 combinations!
there are five places for 2,4,5 so we have 5!/2!3!=10 combinations
must be 3!/ 5!(3-5)! => factorial of -ve is undefined => use 3! multiplied by 2! combined set of combinations 3!*2!=24

so far not clear
2,4,5 has 3! =6 ways to replace its position in abcde
we will have 6 number left accept 2,4,5 to arrange the 2 positions left and those 2 positions must be with different number
so total =6*5 combination
total=6*5*10*6=1800
combining a combination of five numbers with the set of three numbers (calculated I don't understand why) with a combination of three numbers straight?
Here is a step-by-step explanation of the approach being used by others:

There are 5 positions to be filled in the integer. A combination of 3 of these 5 positions must be filled by 2, 4, and 5. The number of combinations of 3 that can be made from 5 choices = 5C3 = 10.

Within this combination of 3 positions, we need to count the number of ways to arrange the digits 2, 4 and 5. The number of ways to arrange 3 elements is 3! = 6.

There are 2 positions left. These 2 positions can be filled by any of the remaining 6 digits. The number of ways to arrange 2 of 6 elements = 6!/(6-2)! = 30.

To form our 5-digit integer, we multiply the results above: 10*6*30 = 1800.

Hope this helps!
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by Night reader » Fri Dec 10, 2010 7:44 am
thanks Mitch I see the logic now.
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