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# A company has assigned a distinct 3-digit code number to each of its 3

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Math Expert
Joined: 02 Sep 2009
Posts: 52344
A company has assigned a distinct 3-digit code number to each of its 3  [#permalink]

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28 Oct 2018, 23:09
00:00

Difficulty:

25% (medium)

Question Stats:

77% (01:06) correct 23% (01:58) wrong based on 53 sessions

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A company has assigned a distinct 3-digit code number to each of its 330 employees. Each code number was formed from the digits 2, 3, 4, 5, 6, 7, 8, 9 and no digit appears more than once in any one code number. How many unassigned code numbers are there?

A. 6
B. 58
C. 174
D. 182
E. 399

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Re: A company has assigned a distinct 3-digit code number to each of its 3  [#permalink]

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28 Oct 2018, 23:29
1
Total of 8 distinct digits.
Number of distinct 3 digit numbers that can be formed are 8*7*6 = 336.

Number of employees = 330.
6 code numbers are unassigned.

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Re: A company has assigned a distinct 3-digit code number to each of its 3  [#permalink]

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28 Oct 2018, 23:30
Total no. of codes possible is $$8*7*6 = 336$$

Since it's a three digit code, the first number can be selected in 8 ways from the given numbers. As no digit repeats in a code, the second digit can be selected in 7 ways and the third digit in 6 ways. So a total of 336 codes are possible.

Now for 330 employees, the codes are assigned and the remaining 6 are unused.

OPTION : $$A$$
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Re: A company has assigned a distinct 3-digit code number to each of its 3  [#permalink]

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30 Oct 2018, 02:53

Solution

Given:
• Number of employees = 330 = number of 3-digit codes assigned
• Each code was formed from the digits, {2, 3, 4, 5, 6, 7, 8, 9}
• No digit appears more than once in any code number

To find:
• The number of unassigned code numbers

Approach and Working:
• Total number of 3-digit code numbers formed using the digits, {2, 3, 4, 5, 6, 7, 8, 9} without any repetition are 8 * 7 * 6 = 336
• Therefore, number of unassigned codes are 336 – 330 = 6

Hence, the correct answer is Option A

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Re: A company has assigned a distinct 3-digit code number to each of its 3 &nbs [#permalink] 30 Oct 2018, 02:53
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