# Moderate Permutation-Combination Solved QuestionAptitude Discussion

 Q. A local delivery company has three packages to deliver to three different homes. If the packages are delivered at random to the three houses, how many ways are there for at least one house to get the wrong package?
 ✖ A. $3$ ✔ B. $5$ ✖ C. $3!$ ✖ D. $5!$

Solution:
Option(B) is correct

The possible outcomes that satisfy the condition of "at least one house gets the wrong package" are:

One house gets the wrong package or two houses get the wrong package or three houses get the wrong package.

We can calculate each of these cases and then add them together, or approach this problem from a different angle.

The only case which is left out of the condition is the case where no wrong packages are delivered.

If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above.

There is only one way for no wrong packages delivered to occur.

This is the same as everyone gets the right package.

The first person must get the correct package and the second person must get the correct package and the third person must get the correct package.

⇒ $1×1×1 = 1$

Determine the total number of ways the three packages can be delivered.

⇒ $3×2×1 = 6$

The number of ways at least one house gets the wrong package is:

⇒ $6 - 1 = 5$

Therefore there are \textbf{5} ways for at least one house to get the wrong package.

## (1) Comment(s)

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Indirect approach easier:

Total outcomes : 3!

Chances for atleast one wrong = total outcomes - the outcome that they're all right (only 1 way to do it right)

6 - 1 = 5