How many three-digit numerals begin With a digit that represents a pri

dave13 if I may chip in my two cents,

You have to use factorial to list out the possibke permutations in an order.

Here you are just selecting numbers ( digits of a three digit number with constraints)

So your answer is almost correct albeit the factorial sign.

4 ways to choose a prime.

The three digit number will begin and end with a prime and the middle digit will have 10 options.

Hence, 4*10*4 = 160

Does this make sense?

Best,
Gladi

No ways to select the first no is 4
no of ways to select middle no is 10
Why is the no ways to select the third no is 4 and not 3? If we select 4, are not we repeating same combination?
Please clarify.

yes we are repeating it,
but thats okay, according to the condition...
Start with a prime digit, and end with a prime digit, doesn't say, "it can't have repeating prime digits"
then, the answer will be 4*10*3 = 120
I guess so

here the answer is fine, 4*10*4 = 160.

Take the task of creating three-digit numerals and break it into stages.

Stage 1: Select a digit for the hundreds position
This digit can be 2, 3, 5, or 7
So, we can complete stage 1 in 4 ways

Stage 2: Select a digit for the tens position
This digit can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9
So we can complete the stage in in 10 ways

Stage 3: Select a digit for the units position
This digit can be 2, 3, 5, or 7
So, we can complete this stage in 4 ways

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create three-digit numerals) in (4)(10)(4) ways (= 160 ways)

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Solution:

Since 2, 3, 5, and 7 are the only single digit primes, there are 4 choices for the hundreds digit and 4 choices for the units digit. However, since there is no restriction for the tens digit, there are 10 choices. Therefore, there are 4 x 10 x 4 = 160 such 3-digit numbers.

Answer: C

Had to use recognize a pattern to figure this one out.
The only hundreds digits that would qualify are 2, 3, 5, 7.
202, 212, 222, 232, 242, 252, 262, 272, 282, 292. <----10 digits. These digits can also end in 3, 5, or 7. In total, there are 40 digits in the two-hundred series that satisfies the condition.

Repeat the same thing for the other three series.
40 x 4 = 160

Choices for the 1st position = {2, 3, 5, 7} = 4
Choices for the 2nd position = {0, 1, 2, ... 10} = 10
Choices for the 3rd position = {2, 3, 5, 7} = 4

Total numbers possible = 4*10*4 = 160

Option C.

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