Bunuel wrote:

A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digits 6?

A. \(\frac{860}{90,000}\)

B. \(\frac{810}{100,000}\)

C. \(\frac{858}{100,000}\)

D. \(\frac{860}{100,000}\)

E. \(\frac{1530}{100,000}\)

We're going to apply the MISSISSIPPI rule at the end of my solution, so here's what it says:

When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....] So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:

There are

11 letters in total

There are

4 identical I's

There are

4 identical S's

There are

2 identical P's

So, the total number of possible arrangements =

11!/[(

4!)(

4!)(

2!)]

-----NOW ONTO THE QUESTION------------------------------

We'll consider two cases:

case i: We have three 6's and two DIFFERENT digits (e.g., 66612)

case ii: We have three 6's and two IDENTICAL digits (e.g., 66677)

case i: We have three 6's and two DIFFERENT digits (e.g., 66612)We already have three 6's. So, we must select 2 different digits from (0,1,2,3,4,5,7,8 and 9)

We can do this in 9C2 ways (=

36 ways)

Now that we've selected our 5 digits, we must ARRANGE them, which means we can use the MISSISSIPPI rule.

We can arrange 3 identical digits and 2 different digits in 5!/3! ways =

20 ways

So, we can have three 6's and two DIFFERENT digits in (

36)(

20) ways (=

720 ways)

case ii: We have three 6's and two IDENTICAL digits (e.g., 66677)We already have three 6's. So, we must select 1 digit (which we'll duplicate)

Since we're selecting 1 digit from (0,1,2,3,4,5,7,8,9) we can do so in

9 ways

Now that we've selected our 5 digits, we must ARRANGE them, which means we can use the MISSISSIPPI rule.

We can arrange 3 identical 6's and 2 other identical digits in 5!/3!2! ways =

10 ways

So, we can have three 6's and two IDENTICAL digits in (

9)(

10) ways (=

90 ways)

So, TOTAL number of ways to have three 6's =

720 +

90 =

810Since there are 100,000 possible 5-digit codes, P(having exactly three 6's) = 810/100,000

Answer: B

Cheers,

Brent

_________________

Test confidently with gmatprepnow.com