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The security gate at a storage facility requires a fivE [#permalink]
23 Feb 2011, 21:05

00:00

A

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D

E

Difficulty:

25% (medium)

Question Stats:

64% (02:11) correct
36% (01:10) wrong based on 101 sessions

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?

Re: 5-digit security code [#permalink]
24 Feb 2011, 03:16

1

This post received KUDOS

Expert's post

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

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.

Re: The security gate at a storage facility requires a fivE [#permalink]
24 Apr 2013, 22:13

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

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