All categories

2 years ago

How many different four lettered codes can be formed if the first letter must be an S or a T

Mathematics

1 answer:

emmainna [20.7K]2 years ago

8 0

Answer:

27,600

Step-by-step explanation:

We use the following formula:-

Number of permutations of r from n

= nPr

= n! / (n-r)!

I am assuming you don't allow repeated letters in the 4 letter codes.

Take S as the first letter. Then we have 3 letters extra out of 25 letters.

Using the above formula the possible ways of doing this is

25! / (25 - 3)!

25! / 22!

= 25*24*23

= 13800.

The same number is obtained if the first letter is a T.

So the answer is 2*13800

= 27,600

You might be interested in

The measure of one angle is six more than half the measure of another angle. If the angles are complementary, find both measures

Snowcat [4.5K]

Answer:

The measures of the two angles are 34° and 56°

Step-by-step explanation:

* Lets explain how to solve the problem

- The complementary angles are the angles whose the sum

of their measures is 90°

- The measure of one angle is six more than half the measure

of another angle

- Let the measure of the second angle angle is x

∵ The measure of the 2nd angle is x

∵ The measure of the first angle is 6 more than the half of the

measure of the second angle

∴ The 1st angle is 6 + 1/2 x

∵ The two angles are complementary

∴ The sum of their measures is 90°

∴ x + (6 + 1/2 x) = 90

∴ (x + 1/2 x) + 6 = 90

∴ 3/2 x + 6 = 90

- Subtract 6 from both sides

∴ 3/2 x = 84

- Multiply both sides by

∴ 3x = 168

- Divide both sides by 3

∴ x = 56

∵ x is the measure of the 2nd angle

∴ The measure of the 2nd angle is 56°

∵ The measure of the 1st angle = 6 + 1/2 x

∵ x = 56

∴ The measure of the 1st angle = 6 + 1/2(56) = 34°

* The measures of the two angles are 34° and 56°

4 0

2 years ago

In terms of the number of moose, what is the relative frequency of male moose, female moose, adult moose, and baby moose? Write

user100 [1]

Answer:

I'm sorry is there a chart/graph, etc. to give us more information?

Without the added information there is no way for me to answer.

If you repost this question with a picture I would be more than happy to help.

6 0

3 years ago

Read 2 more answers

What is the value of -16?

timama [110]

Answer:

the answer would be -4

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

How many solutions does the system have?

serious [3.7K]

The system has two solutions.

You can tell because the second equation is a quadratic and the first is a line, so the line passes through the quadratic twice.

8 0

2 years ago

7) PG & E have 12 linemen working Tuesdays in Placer County. They work in groups of 8. How many

BabaBlast [244]

Part A

Since order matters, we use the nPr permutation formula

We use n = 12 and r = 8

$_{n}P_{r} = \frac{n!}{(n-r)!}\\\\_{12}P_{8} = \frac{12!}{(12-8)!}\\\\_{12}P_{8} = \frac{12!}{4!}\\\\_{12}P_{8} = \frac{12*11*10*9*8*7*6*5*4*3*2*1}{4*3*2*1}\\\\_{12}P_{8} = \frac{479,001,600}{24}\\\\_{12}P_{8} = 19,958,400\\\\$

There are a little under 20 million different permutations.

<h3>Answer: 19,958,400</h3>

Side note: your teacher may not want you to type in the commas

============================================================

Part B

In this case, order doesn't matter. We could use the nCr combination formula like so.

$_{n}C_{r} = \frac{n!}{r!(n-r)!}\\\\_{12}C_{8} = \frac{12!}{8!(12-8)!}\\\\_{12}C_{8} = \frac{12!}{4!}\\\\_{12}C_{8} = \frac{12*11*10*9*8!}{8!*4!}\\\\_{12}C_{8} = \frac{12*11*10*9}{4!} \ \text{ ... pair of 8! terms cancel}\\\\_{12}C_{8} = \frac{12*11*10*9}{4*3*2*1}\\\\_{12}C_{8} = \frac{11880}{24}\\\\_{12}C_{8} = 495\\\\$

We have a much smaller number compared to last time because order isn't important. Consider a group of 3 people {A,B,C} and this group is identical to {C,B,A}. This idea applies to groups of any number.

-----------------

Another way we can compute the answer is to use the result from part A.

Recall that:

nCr = (nPr)/(r!)

If we know the permutation value, we simply divide by r! to get the combination value. In this case, we divide by r! = 8! = 8*7*6*5*4*3*2*1 = 40,320

So,

$_{n}C_{r} = \frac{_{n}P_{r}}{r!}\\\\_{12}C_{8} = \frac{_{12}P_{8}}{8!}\\\\_{12}C_{8} = \frac{19,958,400}{40,320}\\\\_{12}C_{8} = 495\\\\$

Not only is this shortcut fairly handy, but it's also interesting to see how the concepts of combinations and permutations connect to one another.

-----------------

<h3>Answer: 495</h3>

5 0

2 years ago