Evaluate each expression. 6!= 3! - 21 = 6!/3!=

Mathematics

2 answers:

saw5 [17]3 years ago

5 0

Answer:

6! = 720

3! x 2! = 12

6!/3! = 120

Step-by-step explanation:

Margarita [4]3 years ago

3 0

Answer:

Step-by-step explanation:

6! = 6*5*4*3*2*1 =720

3! - 21 = 3*2*1 - 21 = 6 - 21 = (-15)

$\frac{6!}{3!}=\frac{6*5*4*3*2*1}{3*2*1}\\\\=6*5*4=120$

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Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any tw

PolarNik [594]

Answer:

Using backward reasoning we want to show that "We can never get nine 0's".

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

There were 9 0's:

We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

There were 9 1's:

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.