Consider the name as MARIAHOSSAIN and Roll number as 21101164

Question

1.Find the total number of arrangements from the letters of your name taken all at a time without changing the the positions of vowels and consonants.

2.find the total number of three digit numbers possible from the digits of your roll number.

3.Consider a license plate of 8 letters long.First 4 letters include only alphabets and the second 4 letters include digits only. How many license plate can be made in this way if the Alphabet comes from your name and digits come from your roll number?

Answer 1

I can help!

1) For question 1, if you want to find the possible number of arrangements you can make with MARIAHOSSAIN without changing the order of vowels and consonants, first count the number of vowels and consonants.

We can see that there are 6 vowels (A,I,A,O,A,I) and 6 consonants (M,R,H,S,S,N)

Therefore we do this simple equation: 6!x6! or in other words: (6x5x4x3x2x1) x (6x5x4x3x2x1)

This gives us the answer 518,400.

2) This time, you want to keep the position of vowels and consonants in MARIAHOSSAIN while finding the possible number of arrangements. In the last question, we counted 6 vowels in the word MARIAHOSSAIN, and now we must find which vowels are alike.

Within the vowels that we found, we can see that there are 3 A's and 2 I's.

Therefore, the equation would look something like this: 6!/(3!x2!). I will break this down below:

The total number of arrangements of vowels we found in the last equation is 6, and this number

will be on top of the fraction. On the bottom of the fraction are the types that are alike, so the 3!

represents the 3 A's that are alike and the 2! represents the 2 I's in the word MARIAHOSSAIN.

We can conclude that there are 6!/(3!x2!)=60 ways to arrange vowels

Now let's do the same thing for the consonants.

In the last question, we counted 6 consonants, and we can see that there are 2 S's.

Therefore, the equation would look something like this: 6!/2! which equals 360, so there are 360

ways to arrange the consonants.

Now we just multiple the number of ways to arrange vowels (60) with the number of ways to

arrange consonants (360) which will be: 60x360= 21,600

Therefore, the answer to this question will be 21,600

3) I believe the answer to this question depends on whether repetition of numbers is allowed, but

since it is not explicitly mentioned in the question, we will allow for them to happen. However, I will

include the answer for without repeats as well just in case.

We know that the first number in the three digits CANNOT be 0 because that would not make a three

digit number (for example 012 or 12 is not a three digit number). Therefore, excluding 0, we have 4

possible choices for the first number of our three digit number given our roll number (1,2,4, or 6).

The second and third number in the three digit number CAN be 0, so we have 5 possible choices that they could be given our roll number (0,1,2,4, or 6).

Therefore, since our first digit has 4 possible choices, our second digit has 5 possible choices, and our third digit has 5 possible choices, then our equation would look like this:

4 x 5 x 5 which equals 100. So if the digits can be repeated in our three digit number, the answer would be 100.

***In the case that the digits cannot be repeated, we have to solve this a different way.

The first digit still has 4 possible choices which include 1,2,4, and 6 and exclude 0.

This time, the second digit only has 4 possible choices. Why? This is because the first number HAS to be either 1,2,4 or 6. And since numbers cannot be repeated, the second digit loses one of its possible choices from that list of numbers.

To give an example, if the first digit is 1, the second digit can no longer be 1 because it is taken and numbers cannot repeat. Therefore its only possible choices would be 0, 2, 4 or 6 which is only 4 choices.

Finally, the third digit now only has 3 possible choices using the same thought process as mentioned for the second digit. The first and second digit have taken two of the possible choices that the third digit could be, so now it only has 3 possible choices.

So the equation for non-repeating numbers would be: 4 x 4 x 3 which equals 48.

(Please let me know if this is still wrong and I will try to help you solve it another way)

4) It seems like there is no issue with repeating numbers or letters for this question, so I will provide

the answer for a scenario where repeating numbers and letters are allowed! There also seems to be

no requirement to keep the letter or numbers in position or in a certain order so I will exclude that also.

Since the first 4 letters of the license plate only include alphabets from your name, we have these 8

choices: M, A, R, I, H, O, S, and N. Since repetition of these letters are allowed and don't have to be

kept in any sort of order or position, all of the letters have an equal chance to be any of the four letters on the license plate. Therefore, our equation would be: 8 x 8 x 8 x 8 which equals 4096.

The last 4 letters include digits only and come from our roll numbers which include 0,1,2,4, and 6. Without worrying about repetition, we can conclude that each digit has 5 possibilities.

Therefore, our equation would be: 5 x 5 x 5 x 5 which equals 625.

Now let's multiply our totals together: 4096 x 625 = 2,560,000

This means that 2560000 license plate combinations can be made!