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A company plans to assign identification numbers to its empl [#permalink]
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A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible? (A) 3,024 (B) 4,536 (C) 5,040 (D) 9,000 (E) 10,000
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Originally posted by judokan on 23 Aug 2008, 08:21.
Last edited by Bunuel on 26 Feb 2013, 03:20, edited 1 time in total.
Edited the question and added the OA.



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A company plans to assign identification numbers to its empl [#permalink]
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23 Aug 2008, 08:34
judokan wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
3024 4536 5040 9000 10000 = No. of ways select first digit (other than 0) * No of wasy select second digit (exclude first digit selected) * no of ways select 3rd digit (exclude first 2) * no of ways to select 4 th digit (excllude first 3 digits) = 9*9*8*7= 4536
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Re: A company plans to assign identification numbers to its employees. Eac [#permalink]
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09 Mar 2011, 22:20
geisends wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
A)3,024 B)4,536 C)5,040 D)9,000 E)10,000 First digit (1000th place) can be any of 0 to 9 except 0, so it can be chosen in nine ways Second digit (100th place) can be any of 0 to 9 except the one already chosen for 1000th place, so it can be chosen in nine ways Third digit (10th place) can be any of 0 to 9 except the ones already chosen for 1000th place and 100th place, so it can be chosen in eight ways Fourth digit (units place) can be any of 0 to 9 except the ones already chosen for 1000th place, 100th place and 10th place, so it can be chosen in seven ways Total number of ways = 9*9*8*7 = 81*56. Only option with 6 in units place is B, so answer is B.



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A company plans to assign identification numbers to its empl [#permalink]
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26 Feb 2013, 03:27
judokan wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
(A) 3,024 (B) 4,536 (C) 5,040 (D) 9,000 (E) 10,000 Notice that we are told that "Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0". The first digit can take 9 values from 1 to 9 inclusive; The second digit can also take 9 values (9 digits minus the one we used for the first digit plus 0); The third digit can take 8 values; The fourth digit can take 7 values. Total = 9*9*8*7 = something with the units digit of 6. Answer: B. Hope it's clear.
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Re: A company plans to assign identification numbers to its empl [#permalink]
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23 Mar 2013, 08:28
Bunuel wrote: judokan wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
(A) 3,024 (B) 4,536 (C) 5,040 (D) 9,000 (E) 10,000 The first digit can take 9 values from 1 to 9 inclusive; The second digit can also take 9 values (9 digits minus the one we used for the first digit plus 0); The third digit can take 8 values; The fourth digit can take 7 values. Total = 9*9*8*7 = something with the units digit if 6. Answer: B. Hope it's clear. Hi Bunnel, I took 9*10*10*10=9000 i thought he didn't mention any thing such as repetition not allowed or not allowed may i know if i misread the question? May i know where i went wrong?
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A company plans to assign identification numbers to its empl [#permalink]
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23 Mar 2013, 08:52
mydreammba wrote: Bunuel wrote: judokan wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
(A) 3,024 (B) 4,536 (C) 5,040 (D) 9,000 (E) 10,000 The first digit can take 9 values from 1 to 9 inclusive; The second digit can also take 9 values (9 digits minus the one we used for the first digit plus 0); The third digit can take 8 values; The fourth digit can take 7 values. Total = 9*9*8*7 = something with the units digit if 6. Answer: B. Hope it's clear. Hi Bunnel, I took 9*10*10*10=9000 i thought he didn't mention any thing such as repetition not allowed or not allowed may i know if i misread the question? May i know where i went wrong? "Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0"
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Re: A company plans to assign identification numbers to its empl [#permalink]
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31 Mar 2014, 08:32
Four DIFFERENT digits. 9*9*8*7.
Only answer with units digit 6 is B.
Hope this helps Cheers J



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Re: A company plans to assign identification numbers to its empl [#permalink]
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08 Apr 2016, 10:59
Bunuel wrote: judokan wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
(A) 3,024 (B) 4,536 (C) 5,040 (D) 9,000 (E) 10,000 The first digit can take 9 values from 1 to 9 inclusive; The second digit can also take 9 values (9 digits minus the one we used for the first digit plus 0); The third digit can take 8 values; The fourth digit can take 7 values. Total = 9*9*8*7 = something with the units digit of 6. Answer: B. Hope it's clear. I got the same answer almost but I multiplied it by 4! in the end assuming we could rearrange the digits and get a different number. Can you please explain why this is not correct?? Thank you.



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Re: A company plans to assign identification numbers to its empl [#permalink]
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09 Apr 2016, 01:47
MeghaP wrote: Bunuel wrote: judokan wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
(A) 3,024 (B) 4,536 (C) 5,040 (D) 9,000 (E) 10,000 The first digit can take 9 values from 1 to 9 inclusive; The second digit can also take 9 values (9 digits minus the one we used for the first digit plus 0); The third digit can take 8 values; The fourth digit can take 7 values. Total = 9*9*8*7 = something with the units digit of 6. Answer: B. Hope it's clear. I got the same answer almost but I multiplied it by 4! in the end assuming we could rearrange the digits and get a different number. Can you please explain why this is not correct?? Thank you. The method used already takes care of all different arrangements. Try to test with smaller numbers to check.
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Re: A company plans to assign identification numbers to its empl [#permalink]
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21 Apr 2017, 23:42
judokan wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
(A) 3,024 (B) 4,536 (C) 5,040 (D) 9,000 (E) 10,000 The first thing to note is that the possibility of choices is 09 which actually is ten numbers if you count [0,1,2,3,4,5,6,7,8,9] yet the first number of the digit cannot have 0 so there is a pool of 9 choices to choose from for the first number 9 x 9 x 8 x 7 = 4536 Thus B



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Re: A company plans to assign identification numbers to its empl [#permalink]
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11 May 2017, 21:40
1st digit can be anything from 1 to 9, So 9 possible digits 2nd can be anything from 0 to 9 but not the one used as 1st digit, so again 10 1 = 9 possible values 3rd can be anything from 0 to 9 except the two digits used as 1st & 2nd digit, so 102 = 8 possible values 4th can be anything from 0 to 9 except the 3 digits used as 1st, 2nd and 3rd digit, so 103 = 7 Hence total possibilities = 9x9x8x7 = 4536 Hit kudos if you like the solution!!
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Re: A company plans to assign identification numbers to its empl [#permalink]
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11 May 2017, 22:48
judokan wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
(A) 3,024 (B) 4,536 (C) 5,040 (D) 9,000 (E) 10,000 Since we have to form 4 digit number with 1st digit nonzero and all different digits. It can be formed in following way 1st digit > 9 ways (1 to 9) 2nd digit > 9 ways (0 to 9 excluding 1st digit) 3rd digit > 8 ways ( 0 to 9 excluding 1st and 2nd digit) 4th digit > 7 ways ( 0 to 9 excluding 1st , 2nd and 3rd digit) So total no. of ways = 9*9*8*7 = 4536 ways So different identification numbers possible : 4536
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Re: A company plans to assign identification numbers to its empl [#permalink]
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20 May 2017, 20:25
judokan wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
(A) 3,024 (B) 4,536 (C) 5,040 (D) 9,000 (E) 10,000 1. Ordering is important and there is no repetition, so this is an nPr problem 2. Is there a constraint?. The first digit cannot be zero 3. Number of ways the first digit can be selected is 9 4. Number of ways the second third and last digits can be ordered is 9P3 5. Total number of permutations is 9*9*8*7=4536
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A company plans to assign identification numbers to its empl [#permalink]
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07 Sep 2017, 06:42
Bunuel wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
(A) 3,024 (B) 4,536 (C) 5,040 (D) 9,000 (E) 10,000
The first digit can take 9 values from 1 to 9 inclusive; The second digit can also take 9 values (9 digits minus the one we used for the first digit plus 0); The third digit can take 8 values; The fourth digit can take 7 values.
Total = 9*9*8*7 = something with the units digit if 6.
Answer: B.
Hope it's clear. Let's say the question would have said that the first number could as well be a zero, then the number of identification numbers would be : 10*9*8*7 ? Am i going right?



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Re: A company plans to assign identification numbers to its empl [#permalink]
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07 Sep 2017, 06:48
SinhaS wrote: Bunuel wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
(A) 3,024 (B) 4,536 (C) 5,040 (D) 9,000 (E) 10,000
The first digit can take 9 values from 1 to 9 inclusive; The second digit can also take 9 values (9 digits minus the one we used for the first digit plus 0); The third digit can take 8 values; The fourth digit can take 7 values.
Total = 9*9*8*7 = something with the units digit if 6.
Answer: B.
Hope it's clear. Let's say the question would have said that the first number could as well be a zero, then the number of identification numbers would be : 10*9*8*7 ? Am i going right? __________________ Yes, that's correct.
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Re: A company plans to assign identification numbers to its empl [#permalink]
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11 Sep 2017, 11:41
judokan wrote: A company plans to assign identification numbers to its employees. Each number is to consist of four different digits from 0 to 9, inclusive, except that the first digit cannot be 0. How many different identification numbers are possible?
(A) 3,024 (B) 4,536 (C) 5,040 (D) 9,000 (E) 10,000 There are 9 choices for the first digit (1 through 9, inclusive). The second digit can be any of the 10 digits (0 through 9, inclusive) EXCEPT it can’t repeat the first digit; thus, there are 9 options for the second digit. The third digit can’t repeat either of the first two digits, so there are 8 options. Similarly, the fourth digit can’t repeat any of the first 3 digits, so there are 7 options. Thus, the total number of options is 9 x 9 x 8 x 7 = 4,536. Answer: B
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