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The security gate at a storage facility requires a fivE [#permalink]

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23 Feb 2011, 22:05

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The security gate at a storage facility requires a five-digit lock code. If the lock code must consist only of digits from 1 through 7, inclusive, with no repeated digits, and the first and last digits of the code must be odd, how many lock codes are possible?

The security gate at a storage facility requires a five-digit lock code. If the lock code must consist only of digits from 1 through 7, inclusive, with no repeated digits, and the first and last digits of the code must be odd, how many lock codes are possible?

a)120 b)240 c)360 d)720 e)1440

X-X-X-X-X --> there are 4 odd digits from 1 to 7, inclusive thus options for the first and the last X's are: 4-X-X-X-3. Other X's can take following values: 4-5-4-3-3 --> 4*5*4*3*3=720.

The security gate at a storage facility requires a five-digit lock code. If the lock code must consist only of digits from 1 through 7, inclusive, with no repeated digits, and the first and last digits of the code must be odd, how many lock codes are possible?

a)120 b)240 c)360 d)720 e)1440

X-X-X-X-X --> there are 4 odd digits from 1 to 7, inclusive thus options for the first and the last X's are: 4-X-X-X-3. Other X's can take following values: 4-5-4-3-3 --> 4*5*4*3*3=720.

Answer: D.

Hi bunuel,

I have a small doubt , why not we are considering that the first and last digit can occur in 2 ways and the middle 3 digits can occur in 6 ways.

for example middle 3 digits can be arranged in 3! ways among themselves after selecting 5c1 4c1 3c1 .

My question may be stupid, please correct my doubt.

The security gate at a storage facility requires a five-digit lock code. If the lock code must consist only of digits from 1 through 7, inclusive, with no repeated digits, and the first and last digits of the code must be odd, how many lock codes are possible?

a)120 b)240 c)360 d)720 e)1440

X-X-X-X-X --> there are 4 odd digits from 1 to 7, inclusive thus options for the first and the last X's are: 4-X-X-X-3. Other X's can take following values: 4-5-4-3-3 --> 4*5*4*3*3=720.

Answer: D.

Hi bunuel,

I have a small doubt , why not we are considering that the first and last digit can occur in 2 ways and the middle 3 digits can occur in 6 ways.

for example middle 3 digits can be arranged in 3! ways among themselves after selecting 5c1 4c1 3c1 .

My question may be stupid, please correct my doubt.

Thanks

Consider this: two digit code XX, each digit must be distinct and can be 1, 2 or 3.

First digit can take 3 values (1, 2, or 3) and the second can take 2 values, total 3*2=6 codes: 12 13 21 23 31 32.

Similarly, for the original question: the first digit can take 4 values and the last digit can take 3 values, total 4*3. The same way for the middle 3 digits: the second digit can take 5 values (7 minus two we already used for first and last), the third 4 and the fourth 3.

The security gate at a storage facility requires a five-digit lock code. If the lock code must consist only of digits from 1 through 7, inclusive, with no repeated digits, and the first and last digits of the code must be odd, how many lock codes are possible?

A. 120 B. 240 C. 360 D. 720 E. 1440

1. There is one large group i.e., digits 1 through 7, from which 5 digits should be selected. 2. The first and the last digits have a constraint that they should be odd. There is also a general constraint that each digit should be different. 3. Considering the above constraints the first and the last digits can be selected in \(4P2\) ways. 4 because there are 4 odd digits out of which they can be selected and because there should be no repetition it is \(4P2\) ways for those 2 positions. 4. The second digit cannot have the same digit as the first and the last digit. Therefore it can be chosen out of 7-2=5 digits. Similarly the third digit cannot have the same digit as the first, second and last. So it can be chosen in 4 ways and the fourth digit similarly in 3 ways. 5. So the number of lock codes possible is \(4P2 * 5* 4*3 = 720\)

Re: The security gate at a storage facility requires a fivE [#permalink]

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06 May 2015, 13:27

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Re: The security gate at a storage facility requires a fivE [#permalink]

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02 Jul 2016, 16:52

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Re: The security gate at a storage facility requires a fivE [#permalink]

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05 Oct 2016, 07:04

Hi Bunuel, Small clarification is required on this question.

Lets say.. First and last digits should be odd numbers . so it can be done in 5 ways (First) & 4 Ways (Last). Now, Since 5 digits are remaining for 3 places, ( 5C3 X 3!) can be done.... Finally answer is 360 ( can you explain?)

Re: The security gate at a storage facility requires a fivE [#permalink]

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09 Oct 2016, 13:00

rajendra00 wrote:

Hi Bunuel, Small clarification is required on this question.

Lets say.. First and last digits should be odd numbers . so it can be done in 5 ways (First) & 4 Ways (Last). Now, Since 5 digits are remaining for 3 places, ( 5C3 X 3!) can be done.... Finally answer is 360 ( can you explain?)

Concentration: General Management, Entrepreneurship

GPA: 3.8

WE: Engineering (Energy and Utilities)

The security gate at a storage facility requires a fivE [#permalink]

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03 Jul 2017, 06:05

IndigoIntentions wrote:

The security gate at a storage facility requires a five-digit lock code. If the lock code must consist only of digits from 1 through 7, inclusive, with no repeated digits, and the first and last digits of the code must be odd, how many lock codes are possible?

A. 120 B. 240 C. 360 D. 720 E. 1440

Security gate requires a five digit lock code. Lock code can take digits from 1 to 7 inclusive i.e. 1,2,3,4,5,6,7... Odd nos. among these nos are 1,3,5,7,total 4 in numbers.

Lets make boxes below to check in how many digits can be placed at the 5 different places of the lock code.

Let me explain this.. At the first place we can place any of the 4 digits. At the last place we can place any of 3 digits. (as digits can't be repeated.)

Now there are 5 digits left. So, at the 2nd place we can place any of 5 digits. at the 3rd place we can place any of 4 digits. at the 4th place we can place any of 3 digits.

So, total number of lock codes possible = 4 * 5 * 4 * 3 * 3 = 80 * 9 = 720.

Concentration: General Management, Entrepreneurship

GPA: 3.8

WE: Engineering (Energy and Utilities)

Re: The security gate at a storage facility requires a fivE [#permalink]

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05 Jul 2017, 06:27

1

This post received KUDOS

ameyaprabhu wrote:

Hi,

I always get confused as to when to use the arrangement formula, i.e. n!, and when to use the combination formula, i.e. nCp.

In the given question I understood the selection of the 1st and last place.

My doubt is for the middle 3 places. Now since we have to 'choose 3 digits out of 5' to fill the middle 3 places, why can't we do 5C3?

Use arrangement formula n! only when u need to arrange the things like.

I am taking a very simple example for this . If u want to make an arrangement of numbers 3,4,5 = 3! = 6 So lets see it by making the arrangement (345),(354),(534),(543)(453),(435) = Total 6 .

nCp is used when u have to make selection . For e.g. u need to select the 2 numbers from the three given numbers 3,4,5, = 3C2 = 3 ways.

Coming to the given question. U have already understood the the first and last numbers,

Now for the middle places, We have to choose 3 digit out of 5 to fill the places. So it can be done in 5 x4 x3 ways = 60 ways

It can also be done in the way you suggested. Lets do it that way. So according to you choose 3 digits out of 5 to fill the middle 3 places = 5C3 = 5!/3!/2! But these 3 numbers which you have selected out of 5 can also be arranged among themselves in 3! ways as explained in the example of 3,4,5. So you need to multiply your section with 3!. So the total ways become 5!/3!/2! * 3! = 5!/2! = 5*4*3 = 60 ways.. So answer comes to same ... either your way or my way.. Its just you need to think completely. Don't forget to appreciate with kudos if you like my answer.

I hope it solves your issue. Practice more questions to understand. With practice you will find that you don't need any formula and you can just calculate either way...
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