pariearth wrote:

Q7. Mac has forgotten his friends 8-digit telephone number but remembers the following:

i) the first 3 digit are either 270 or 279.

ii) the digit 0 occurs exactly three times and the digit 9 occurs exactly one time.

iii) the number was an even number.

If he were to use a trial & error method to reach his friend, what is the minimum number of trials he has to make to be sure to succeed?

a. 1664 b. 1544 c. 832 d. 1280 e. 2000

These questions are not at all similar to real GMAT questions - they're way too complicated, and there aren't any real shortcuts. In this phone number question, we have four separate cases:

Case 1: phone number is 270-XXXX-0

- one of the four X's is a nine, so we have 4 choices for where to put the nine

- then one of the three remaining X's is a zero, so we have 3 choices for where to put the zero

- then we have 8 choices for the two remaining digits (they cannot be 0 or 9)

So we have 4*3*8*8 possibilities

Case 2: phone number is 270-XXXX-E, where E is even and nonzero

- we have 4 choices for E (it is 2, 4, 6, or 8)

- one of the four X's is a nine, so we have 4 choices

- two of the remaining three digits are zero, so we have 3 choices (3C2)

- for the remaining digit we have 8 choices (cannot be 0 or 9)

So we have 4*4*3*8 possibilities

Case 3: phone number is 279-XXXX-0

- we have 4C2 = 6 choices for where to put the remaining zeros

- we have 8 choices for each of the two other digits

So we have 6*8*8 possibilities

Case 4: phone number is 279-XXXX-E where E is even and nonzero

- we have 4 choices for E

- we have 4C3 = 4 choices for where to put the three zeros

- we have 8 choices for the other digit

So we have 4*4*8 possibilities.

Adding the possibilities from each case, we get the answer:

4*3*8*8 + 4*4*3*8 + 6 * 8 * 8 + 4 * 4 * 8 = 1664

I guarantee you will never see a problem anywhere close to as awkward as this on the GMAT - when you have cases in a GMAT counting problem and no shortcut that allows you to bypass the case analysis, you will almost always have only two cases to consider.

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