I am stuck and need help in solving.
A company wants to assign its employees ID numbers that have x digits each. The firm has 800 employees, what is the smallest number of digits the firm can use for each ID number. Repeats are allowed.
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Permutation Problem
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DavidG@VeritasPrep
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Hi, Icpanam.
Just imagine that x = 1. Well, in that case, you could assign any digit between 0 and 9, so you'd have 10 options
If x = 2, the first digit could be anything between 0 and 9, and because repeats are allowed, the second digit, could also be anything between 0 and 9. Now we're up to 10*10 = 100 options.
If x = 3, any of the first 3 digits could be anything between 0 and 9, giving us 10*10*10 = 1000 options. That'll cover all the employees.
Just imagine that x = 1. Well, in that case, you could assign any digit between 0 and 9, so you'd have 10 options
If x = 2, the first digit could be anything between 0 and 9, and because repeats are allowed, the second digit, could also be anything between 0 and 9. Now we're up to 10*10 = 100 options.
If x = 3, any of the first 3 digits could be anything between 0 and 9, giving us 10*10*10 = 1000 options. That'll cover all the employees.
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GMATinsight
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You don't really need to apply any counting techniques here.lcpanam wrote:I am stuck and need help in solving.
A company wants to assign its employees ID numbers that have x digits each. The firm has 800 employees, what is the smallest number of digits the firm can use for each ID number. Repeats are allowed.
With 1 digit, you can assign values from 0 to 9 (10 possibilities)
With 2 digits, you can assign values from 00 to 99 (100 possibilities)
With 3 digits, you can assign values from 000 to 999 (1000 possibilities)
So, 3-digit ID numbers will accommodate the 800 employees.
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
