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An office manager must choose a five-digit lock code for the

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An office manager must choose a five-digit lock code for the [#permalink]

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New post 15 Jun 2010, 17:33
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An office manager must choose a five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of the digits is allowed. How many different lock codes are possible.?


Odd numbers 1,3,5,7,9 so first digit 5 options, last place 4 options ( no repetition).
That leaves us with 7 (of 9) digits to choose for the middle three.

So we have 5 x 4 for the irst and last
and 7 x 6 x 5 for the iddle three
Total is 5 x 7 x 6 x 5 x 4 = 4200.

The book shows 5 x 8 x 7 x 6 x 4= 6720.

I think this is wrong.

Any ideas.
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Re: Combinatorics Problem- someone from MGmat please see this [#permalink]

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New post 15 Jun 2010, 19:25
I think you forgot about 0 as a digit.

If we assume the possible numbers are 0 - 9 we get 5 choices for one end, 4 for the other and we are left with 8 choices. So the middle portion would be given be 8x7x6

So the total probability would be: 5x8x7x6x4 = 6720.

Hope this helps.
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Re: Combinatorics Problem- someone from MGmat please see this [#permalink]

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New post 15 Jun 2010, 19:56
ah- didnt think of that. Short sighted. Then it makes sense. the answer didnt list out the evens and I assumed.
Thanks.
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Re: Combinatorics Problem- someone from MGmat please see this [#permalink]

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New post 28 Nov 2011, 16:54
Can someone please explain to me how you come up with the 8x7x6?...I can't figure it out to save my life. I understand the 5 and 4, but have no clue on the middle ones. HELP!
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Re: Combinatorics Problem- someone from MGmat please see this [#permalink]

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Using slot method:
5c1*_*_*_*4c1
Fro the second slot we have 8 digits. similarly 7c1 and 6c1 for the 3rd and 4th slot.
i e. 5c1*8c1*7c1*6c1*4c1
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MGMATCombinatorics Problem- someone from MGmat please see th [#permalink]

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New post 19 May 2013, 11:07
Problem: An office manager must choose a five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of digits is allowed. How many different lock codes are possible?
MGMAT answer: 5 x 8 x 7 x 6 x 4 = 6720.

However, I think 6720 (the answer given in MGMAT) is wrong. Here's my reasoning. If you have (an) odd number(s) for your 2nd, 3rd and/or 4th digits, you cannot repeat that(those) numbers for the last digit. Therefore, this problem needs to be partitioned into 4 groups.

1) ODD (1st digit); 3 EVENs (2nd, 3rd, 4th digits); ODD (5th digit)
5 x 5x 4 x 3 x 4 = 1200

2) ODD (1st digit); 2 EVEN, 1 ODD (2nd, 3rd, 4th digits); ODD (5th digit)
5 x 5 x 4 x 4 x 3 = 1200

3) ODD (1st digit); 1 EVEN, 2 ODDs (2nd, 3rd, 4th digits); ODD (5th digit)
5 x 5x 4 x 3 x 2 = 600

4) ODD (1st digit); 3 ODDs (2nd, 3rd, 4th digits); ODD (5th digit)
5 x 4 x 3 x 2 x 1 = 120

Then the total # of combinations would be 1200 + 1200 + 600 + 120 = 3120.

Please let me know whether there's a flaw in my logic. Thanks!
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Re: An office manager must choose a five-digit lock code for the [#permalink]

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New post 19 May 2013, 22:14
Even Digits : 0,2,4,6,8
Odd Digits : 1,3,5,7,9

Chances of choosing first odd digit = 5C1
Chances of choosing last odd digit = 4C1 (no repetitions)
Remaining middle 3 digits can be taken from 8 digits = 8C1*7C1*6C1

So totally 5C1*4C1*8C1*7C1*6C1 = 6720
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Re: An office manager must choose a five-digit lock code for the [#permalink]

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New post 20 May 2013, 00:31
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jeeyo wrote:
An office manager must choose a five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of the digits is allowed. How many different lock codes are possible.?


Odd numbers 1,3,5,7,9 so first digit 5 options, last place 4 options ( no repetition).
That leaves us with 7 (of 9) digits to choose for the middle three.

So we have 5 x 4 for the irst and last
and 7 x 6 x 5 for the iddle three
Total is 5 x 7 x 6 x 5 x 4 = 4200.

The book shows 5 x 8 x 7 x 6 x 4= 6720.

I think this is wrong.

Any ideas.


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Re: An office manager must choose a five-digit lock code for the [#permalink]

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New post 20 May 2013, 05:50
Vamshiiitk and Bunuel,

Thanks for your responses. I guess my question comes down to this: based on how the question is worded, can we have a 5-digit lock code such as 97537? Here, though the first and last digits are not repeated, 7 is used twice in the entire combination. How do you guys interpret the problem?

Once again, the problem reads
"Problem: An office manager must choose a five-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of digits is allowed. How many different lock codes are possible?"
(it's on pg. 48 of the 5th edition MGMAT Number Properties)

If a code such as 97537 is allowed, then, yes the solution 5C1*8C1*7C1*6C1*4C1 = 6720 correct.

But if NONE of the numbers can be repeated, not just first and the last digits, then you cannot assume 4C1 for the last digit. If the first four digits are all odd numbers, then I am left with only one choice for the last digit, i.e. 1C1 = 1.
Similarly, if the first four digits had 3 odd numbers, then you would have 2C1 = 2 for the last digit. That's why I had to partition the problem into four cases.

My argument is valid only if NONE of the numbers can be repeated.

Please let me know what you think. Thanks.
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Re: An office manager must choose a five-digit lock code for the [#permalink]

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New post 14 Dec 2015, 06:01
a thread of solutions is upgraded significanlty in terms of its value to me, if i see in it a solution by GMATPrepNow. Thanks. I'm still waiting.
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Combinatorics (Office Manager choose a five-digit lock) [#permalink]

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New post 25 Dec 2017, 15:29
Hi all,

An office manager must choose a fice-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of digits is allowed. How many different lock codes are possible?

The MGMAT instructions say that it is necessary to walk through the following two steps:



1.: Start with the restricted slots.
5*[2nd slot]*[3rd slot]*[4th slot]*4

2.: Fill the remaining slots.
After filling the restricted slots, there are only 8,7 and 6 digits left for the remaining slots.
5*8*7*6*4


Until here, everything pefectly makes sense to me.

But: I don't understand why the first odd slot (5) affects the number of possibilities for the following slots, but the last slot (4) does not, when we started filling the restricted slots at the beginning.

It makes sense when we say the office manager picks the number in the order from left to right, right? But what would change if the office manager started with the not-constrained slots. Would the formula change as follows? 9*8*7*5*4 Thus, the number of possibilites would increase, or did I get anything wrong? :|

If this is true, shouldn't the formula to the question be 5*4*7*6*5, since it is necessary to start with the restrictions?

Hope somebody can help me, this is driving me crazy...

Thanks in advance!
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Re: Combinatorics (Office Manager choose a five-digit lock) [#permalink]

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New post 25 Dec 2017, 16:35
I think it should be this way - first digit and last digit must be odd so 5x4 ways as repetition not allowed

Second , third and fourth can be selected in 8x7x6ways as 2 digits already used in first and last . Therefore others need to picked out of 8 digits (0 included)

So the answer should be 5x8x7x6x4 ways = 6720


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Re: An office manager must choose a five-digit lock code for the [#permalink]

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New post 25 Dec 2017, 19:42
vedodadoo12 wrote:
Hi all,

An office manager must choose a fice-digit lock code for the office door. The first and last digits of the code must be odd, and no repetition of digits is allowed. How many different lock codes are possible?

The MGMAT instructions say that it is necessary to walk through the following two steps:



1.: Start with the restricted slots.
5*[2nd slot]*[3rd slot]*[4th slot]*4

2.: Fill the remaining slots.
After filling the restricted slots, there are only 8,7 and 6 digits left for the remaining slots.
5*8*7*6*4


Until here, everything pefectly makes sense to me.

But: I don't understand why the first odd slot (5) affects the number of possibilities for the following slots, but the last slot (4) does not, when we started filling the restricted slots at the beginning.

It makes sense when we say the office manager picks the number in the order from left to right, right? But what would change if the office manager started with the not-constrained slots. Would the formula change as follows? 9*8*7*5*4 Thus, the number of possibilites would increase, or did I get anything wrong? :|

If this is true, shouldn't the formula to the question be 5*4*7*6*5, since it is necessary to start with the restrictions?

Hope somebody can help me, this is driving me crazy...

Thanks in advance!


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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: An office manager must choose a five-digit lock code for the   [#permalink] 25 Dec 2017, 19:42
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