Since we're dealing with a problem where the order matters and the first two letters are already chosen we need to subtract the number of letters and the number of available slots per group.

We use the permutation formula to find the answer, but before that let's check values.

n = 8

k = 5

Now since there are two letters already chosen we have to deduct two from both the value of n and k.

n = 6

k = 3

Now we can use the permutation formula:

$_{n}P_{k}=\dfrac{n!}{(n-k)!}$

$_{6}P_{3}=\dfrac{6!}{6-3)!}$

$_{6}P_{3}=\dfrac{6!}{3!}$

$_{6}P_{3}=\dfrac{6*5*4*3*2*1}{3*2*1}$

The 3*2*1 cancels out and leaves us with:

$_{6}P_{3}=6*5*4$

$_{6}P_{3}=120$

So there are 120 possible ways to arrange eight letters into groups of five where order matters and the first two letters are already chosen.

6 0

3 years ago

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The cost of a baguette and a cup is 1.45 the cost of 2 baguettes and a cup is 2.10 what is the cost of the cup

olga2289 [7]

Answer:

0.80

Step-by-step explanation:

2.10-1.45=0.65

0.65=1 baugette

1.45-0.65=0.8

0.8=1 cup

5 0

3 years ago