Answer:

1.If we had been given “cirkus”, then the answer would be easy. We have 6 ways to pick the 1 st letter, then 5 ways to pick the 2 nd, 4 ways to pick the 3 rd, etc giving 6!=720 permutations.

2.For any of nine first digits, you can chose from the eight remaining digits for the second number

9*8=72 different two digit numbers

UNLESS

You’re willing to accept numbers like 1.2 as “numbers” for the purposes of your question.

This doubles the number of possible numbers to 144, since there are now two possible locations for the decimal.

What about numbers like .34? Most people would insist that that is a three digit number, 0.34. But some conventions do not write that leading zero. This then would mean there are three valid locations for the decimal, thus tripling the 72 configurations to 216

Now to get more esoteric.

What if you don’t require that this be in base ten?

OK, all bases lower than ten are out, because 9 is a digit.

But you left out the digit 0, so this isn’t a full set of available digits. You could be in, for example, base 12, which has the digits 0,1,2,3,4,5,6,7,8,9,a, and b

Why does this matter? Because 12 base 12 is one twelve and two ones, the quantity represented in base 10 by the number 14. That’s OK, that was one of the quantities we already had in our base ten set. But the number, for example, 89 base twelve is 8 twelves and 9 ones, or the quantity represented in base 10 by 105.

This then means that 89 base twelve is a number which does not appear in the 72 numbers we generated in base ten.

And as there are an infinity of bases larger than base ten, there are thus an infinity of two digit numbers we can generate by using only the nine digits you gave us, with no repetition.

The key lesson here, you have to define the terms.

But you probably meant 72.

3.There are 8 choices for awarding first prize. Then there are 7 choices for awarding second prize. And there are 6 choices for awarding third prize. Therefore, there are: 8 *7 *6 =336 ways.

4.The first digit can be any of eight possibilities. For each first digit, the second digit can be any of eight possibilities.

Therefore, there are 8 times 8 possible combinations; this comes to 64.

This includes 11, 22, 33, 44 etc.

It is possible that your question, which is worded slightly ambiguously, could have meant to insist that there must be two different digits, that those combinations which use the same digit twice are not allowed. It is clear that there are 8 such “twin” combinations, so we can see that 64–8= 56 is the answer.

We could also have reached the same conclusion by seeing that for each of the 8 first digits, there are only 7 allowed second digits, as the matching digit is not allowed. This would be 8*7 possible combinations, which means 56.

It is also possible that the question means to state that no combination in which the same two digits are used is allowed, as an example, 12 and 21 are both using the same digits, order unimportant.

If that’s the case, then for the first digit 1 you have 8 possible second digits, for the first digit 2 you have 7 possible second digits, and so forth, so that by the 8th first digit there is only one possible second digit. So 8+7+6+5+4+3+2+1 which comes to 36 combinations.

But 8 of those combinations, again, are doubled digits, so if you had BOTH these unlikely restrictions, you’d be looking at a mere 36–8=28 allowed pairs.

Since any of these could appear in two different ways, though, we’d be back at 56 actual combinations, it’s just that you could have no more than 28 individual locks while obeying the restriction; each of the 56 combinations, when picked, would eliminate two possibilities. So Fred could pick 12, but now 21 becomes an illegal combination, so George has only 54 choices, and so forth, until the 28th person has only two combinations to pick from.

TL;DR: Question is ambiguously worded and cannot be answered with certainty.

5.There are 5 cubes, each cube has a different color and on each cube the numbers 1-6. Someone throws the cubes. How many combinations are there, in which the set of the numbers that appear on the cubes has exactly 3 objects?

I was thinking: we need 3 different numbers and then 2 numbers that appeared already, so - 6⋅5⋅4⋅1⋅1, then multiply by all the ways to order the colors ... so the answer =5!⋅6⋅5⋅4

Step-by-step explanation:

HOPE IT HELPS