Mathematics, 30.06.2019 23:10, cjhamilton8614

If radio station call letters must begin with either k or w and must include either two or three additional letters, how many different possibilities are there?

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The total different possible combinations of letters for the radio stations with 3 or 4 letters, such that they start with k or w, is 36504.

Further explanation

To answer this question we need to find all possible combination of letters with a length of 3 or 4, regarded that those combinations start with k or w. The reason why we need to find all combinations with a length of 3 or 4 is because we must start with k or w, and then add either 2 or 3 letters.

To properly understand this question, first we need to know how many letters we can deal with, if we are dealing with the English alphabet, then we have a total of 26 letters we can use (from a to z).

Let's start simple with an example to understand the concept. Suppose we wish to compute all possible combination of letters of length 2, such that the first letter is k. The answer to this would be 26 since we are counting words like ka, kb, kc, ..., kz, which sums up to 26 (26 possibilities of the second letter, for just 1 possibility of the first letter).

Now let's add a new letter to our previous example, so suppose we wish to compute all possible combination of letters of length 3, such that the first letter is k. The answer to this would be 26^2 , since we are counting words like kaa, kab, kac, ... kaz, kba, kbb, kbc, ..., kbz, ..., kza, kzb, kzc, ... kzz, which sums up to 26^2 (26 possibilities of the third letter, for every 26 possibilities of the second letter, for just 1 possibility of the first letter).

From the previous example, it's straightforward to see that the total number of combination of letters of length 4 such that the first letter is k, is 26^3 . This gives us all which is necessary to solve our problem.

So, if the first letter is k in our combination, the total number of possible combinations of length 3 and 4 will be 26^2 + 26^3 , which is equal to 18252. In case, the first letter of the combination is w, we have the same result, 18252. Therefor the answer to our question will be 18252 + 18252 = 36504 .

Learn moreOrdered combinations: linkUnordered combinations: linkAnother problem on combinations: linkKeywords

Permutations, possibilities, combinations.

answered: oceaneyez

Total number of possibilities to fill the two or three additional letters is $\fbox{\begin\\\ \bf 36504\\\end{minispace}}$ .

Further explanation:

This problem is based on the concept of fundamental principle of counting.

Fundamental principle of counting: If first event can be done in $m_{1}$ ways, second event can be done in $m_{2}$ ways and similarly the $n^{th}$ event can be done in $m_{n}$ ways then the total number of possibilities for all the events to occur together is calculated as follows:

$\fbox{\begin\\\ \math P=m_{1}\times m_{2}\times ...\times m_{n}\\\end{minispace}}$

It is given that the radio station call letters began with the letter either k or w and there are two or three additional letters.

Two cases are formed in which the first case is additional letters are placed and the second case is additional letters are placed.

First case:

Consider that the radio station call letters has three letters in which first letter is either k or w.

$\fbox{\begin\\\ \\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}$

The total number of letters in the English alphabet system is .

It is given that the first letter is either k or w. So, there are total two ways in which the first section can be filled.

$\fbox{\begin\\\ \fbox{\begin\\\ k\\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\ \text{or}\ \fbox{\begin\\\ w\\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\\\end{minispace}}$

Now the next three sections are blank and any letter can be placed in them. It is not given in the question if the repetition of the letters is allowed or not.

We consider that the repetition of letters is allowed.

As per the above statement it is concluded that the each blank in the next two sections can be filled in ways.

So, as per the concept of fundamental principle of counting the total number of possibilities to fill the two additional letters is calculated as follows:

$\begin{aligned}P_{1}&=2\times26\times26\\&=1352\end{aligned}$

Second case:

Consider that the radio station call letters has four letters in which first letter is either k or w.

$\fbox{\begin\\\ \\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}$

The total number of letters in the English alphabet system is 26.

It is given that the first letter is either k or w. So, there are total two ways in which the first section can be filled

$\fbox{\begin\\\ \fbox{\begin\\\ k\\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\ \text{or}\ \fbox{\begin\\\ w\\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\fbox{\begin\\\ \\\end{minispace}}\\\end{minispace}}$

Now the next three sections are blank and any letter can be placed in them. It is not given in the question if the repetition of the letters is allowed or not.

We consider that the repetition of letters is allowed.

As per the above statement it is concluded that the each blank in the next three sections can be filled in ways.

So, as per the concept of fundamental principle of counting the total number of possibilities to fill the three additional letters is calculated as follows:

$\begin{aligned}P_{2}&=2\times26\times26\times26\\&=35152\end{aligned}$