Answer

Verified

347.1k+ views

**Hint:**At first, we must understand whether to use permutation or combination in this question. Then, we must use the proper concept and find the total number of ways in which the parcel can be sent. We can multiply the number of ways in which each individual parcel can be posted, to find the required number.

**Complete step-by-step solution:**

We are given four parcels and five post offices. We must understand here that for posting one parcel, we just need to select any one of the post offices. We need to find that in how many different ways, we can post these four parcels.

Now, since we have understood the question, we can clearly see that we need to make a selection, not arrangement. Thus, we are very clear that we must use combination in this question, and not permutation.

We know that the selection of r distinct objects from a set of n distinct objects, is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. We also know that ${}^{n}{{C}_{r}}$ is read as the combination of r objects from n different objects.

Here, we can see that for any parcel, we need to select 1 post office from a set of 5. Thus, the number of ways, in which we can post this parcel is ${}^{5}{{C}_{1}}$.

Similarly, for the second parcel, we again need to select 1 post office from a set of 5. Thus, the number of ways, in which we can post this parcel is ${}^{5}{{C}_{1}}$.

Similarly, the number of ways we can post the third and fourth parcel is ${}^{5}{{C}_{1}}$ and ${}^{5}{{C}_{1}}$.

Thus, the total number of ways, in which, we can post these four parcels = ${}^{5}{{C}_{1}}\times {}^{5}{{C}_{1}}\times {}^{5}{{C}_{1}}\times {}^{5}{{C}_{1}}$.

We can write, ${}^{5}{{C}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!}$.

And so, we have ${}^{5}{{C}_{1}}=\dfrac{5!}{1!\times 4!}$. Thus, we get ${}^{5}{{C}_{1}}=5$.

So, the total number of ways, in which, we can post these four parcels = $5\times 5\times 5\times 5$

**Hence, the total number of ways in which we can post these four parcels = 625.**

**Note:**We must understand that permutation is used in cases of arrangement, and combination is cases of selection. We must be very clear that the combination is given as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, and the permutation is given as ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

How many crores make 10 million class 7 maths CBSE

The 3 + 3 times 3 3 + 3 What is the right answer and class 8 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE