Answer:

1. How many 3-digit numbers can be formed using only 1, 2, 3?

Well 3 digit numbers formed by 1,2,3 can be repetitive or non repetitive type

If repetition of any number is allowed:

Then 3³ = 3x3x3 = 27 possibilities

Suppose xyz is 3 digit number then

For x we have 3 possibilities, for y 3 possibilities and for x again 3 possibilities if repetition is allowed so possible numbers = 3x3x3 = 27

Numbers are

111, 112, 113, 121, 122, 123, 131, 132, 133

211, 212, 213, 221, 222, 223, 231, 232, 233

311, 312, 313, 321, 322, 323, 331, 332, 333

Total = 27

If repetition of any number is not allowed:

If repetition is not allowed then

Take example of xyz

Now for x we have 3 possibilities

For y we cannot use the number already used for x so (3–1) = 2 possibilities

For z we cannot use the number already used for x and y so we left with (3–2) = 1 possibility

So total 3 digit number possibilities without repetition is 3x2x1 = 6

Numbers are

123, 132, 213, 231, 312, 321

Total = 6

We can also calculation it through permutation :

nPr = n! / (n-r)!

Now we have to form 3 digit number so r=3

We have 3 numbers available so n=3

So

3P3 = 3!/ (3–3)! = 3x2x1 / 0! = 6 /1 = 6

Or if you have required outcome = available resource as in our case both required number = 3 and available digits = 3

We can use directly 3! = 3x2x1 = 6

2. Is this a permutation or combination? Ask yourself, is order important. If it isn’t, then it’s combination - how many ways can we choose 6 of the 10 people to sit in the chairs? If order is important (as it is here) then it’s a permutation - how many ways can we have 6 of 10 people sit in the chairs?

Once you know it’s a permutation, just plug in the numbers: P(10,6)= 10!(10–6)!=151,200 .

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If you wanted to do it as a combination though, you still could. You’d just have to do it in two parts. Part 1: Choose the 6 of 10 people that can sit: C(10,6)= 10!(10–6)!6! . Part 2: Arrange the 6 chosen people in the 6 seats - there are 6! ways to do this.

Then combine multiplicatively: 10!(10–6)!6!∗6!=10!6!(10–6)!6!=10!(10–6)!=151,200 . Of course, when you put it all together, you can see that it just simplifies down to P(10,6). Or in other words, you can see how a combination effectively calculates a permutation, but then removes the ordering from it.

Hope it will help

Step-by-step explanation:

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