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3 years ago

How many 3 letter codes are possible using a-j without repeats?

1 answer:

5 0

There are 10 letters in the set {a, b, c, d, e, f, g, h, i, j} which is the pool of letters to choose from when making these three letter codes.

We have 10 choices for slot 1
Then 9 choices for slot 2. This is because we can't reuse the choice for slot 1
Then 8 choices for slot 3

Overall, there are 10*9*8 = 90*8 = 720 different permutations

Answer: 720

Note: you can use the nPr permutation formula with n = 10 and r = 3 to get the same answer

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The time spent dancing (minutes) and the amount of calories burned can be modeled by the equation c = 5.5t. Which table of value

pishuonlain [190]

To build the table you just have to give values to t and then calculate the corresponding c a per the model.

If the model is c = 5.5 t this is the table

t    c
   5.5 t

0 5.5(0) = 0

1 5.5(1) = 5.5

2      5.5(2) = 11.0

3 5.5(3) = 16.5

4 5.5(4) = 22.0

5 5.5(5) = 27.5

The viables solutions are all where t is equal or greater than 0. You can even use decimal values for t. You cannot use negative values for t.

5 0

4 years ago

Read 2 more answers

What is a object accelerated if its moving in 30m and stops at 5 seconds

Ivan

6 meter / second square.

8 0

4 years ago

Lauren has a bouquet of flowers. 1/5 of the flowers are red. 3/10 of the flowers are white. The rest are yellow. Lauren’s bouque

ss7ja [257]

Answer:

36 flowers

Step-by-step explanation:

Given

$Red = \frac{1}{5}$

$White = \frac{3}{10}$

$Yellow = 18$

Required

Determine the number of flowers

First, we need to determine the fraction of flowers that are yellow;

$R + W + Y = 1$

$\frac{1}{5} + \frac{3}{10} + Y = 1$

$\frac{2+ 3}{10} + Y = 1$

$\frac{5}{10} + Y = 1$

$Y = 1 - \frac{5}{10}$

$Y = \frac{10 -5}{10}$

$Y = \frac{5}{10}$

$Y = \frac{1}{2}$

Let the number of flowers by N;

The yellow flowers can be represented as;

$18 = \frac{1}{2} * N$

Solve for N

$N = \frac{18 * 2}{1}$

$N = \frac{36}{1}$

$N = 36$

5 0

3 years ago

Help binomial distributions

Ymorist [56]

We have the formula to compute the probability of having exactly k successed over n trials, given a probability p of success (and implicitly a probability 1-p of failure), which is

$P(\text{k successed on n trials}) = \binom{n}{k}p^k(1-p)^{n-k}$

Now, the probability of at least 3 successes is the union of the following event: exactly three successes,exactly four successes and exactly five successes.

We can compute their probability and sum them: $\binom{5}{3}\left(\frac{3}{7}\right)^3\left(\frac{4}{7}\right)^2 + \binom{5}{4}\left(\frac{3}{7}\right)^4\left(\frac{4}{7}\right)^1 + \binom{5}{5}\left(\frac{3}{7}\right)^5 \approx 0.36788$

So, the answer is about 36.79%

3 0

3 years ago

Multiply. Express your answer in simplest form (x^2 -4x+3)(-3x^2+7x-1)

mafiozo [28]

Here the answer. -3x^4 +19x^3-38x^2+25x-3

3 0

3 years ago