A realtor uses a lock box to store the keys to a house that is for sale. The access code for the lock box consists of sixsix digits. The first digit cannot be 22 and the last digit must be oddodd. How many different codes are available? (Note that 0 is considered an even number.)

Question

Jax Riggs

Mathematics

17 August, 05:48

A realtor uses a lock box to store the keys to a house that is for sale. The access code for the lock box consists of sixsix digits. The first digit cannot be 22 and the last digit must be oddodd. How many different codes are available? (Note that 0 is considered an even number.)

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Answers (1)

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Cullen Mcbride · Answer 1

450,000 codes

Step-by-step explanation:

Number of digits in the access code = 6

Conditions:

First digit cannot be 2 Last digit must be odd

Since, there are 10 digits in total from 0 to 9 and first digit cannot be 2, there are 9 ways to fill the place of first digit.

Last digit must be odd. As there are 5 odd digits from 0 to 9, there are 5 ways to fill the last digit.

The central 4 digits can be filled by any of the 10 numbers. So, each of them can be filled in 10 ways.

According to the fundamental rule of counting, the total possible codes would be the product of all the possibilities of individual digits.

Therefore,

Number of possible codes = 9 x 10 x 10 x 10 x 10 x 5 = 450,000 codes

Hence, 450,000 different codes are possible for the lock box