A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:
n!/((n-r)!r!)
In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.
5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5
5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10
5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10
5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5
Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.
Now we add together the combinations
1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.
The answer is 32.
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Note: There is also an equation for permutations which is:
n!/(n-r)!
Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.
We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.
I hope this helps! If you have any questions, let me know :)
178 - (-395)
178 + 395 = 573
the distance between -395 and 178 is 573
hope this helps
Communitive Property of Addition
H is 2t
h + t = 84
2t + t = 84
3t =84
t = 28
Thomas is 28 and huilan is 56
The first choice can be any one of the 8 side dishes.
For each of these . . .
The 2nd choice can be any one of the remaining 7.
Total number of ways to pick 2 out of 8 = (8 x 7) = 56 ways .
BUT ...
That doesn't mean you can get 56 different sets of 2 side dishes.
For each different pair, there are 2 ways to choose them . . .
(first A then B), and (first B then A). Either way, you wind up with (A and B).
So yes, there are 56 different 'WAYS' to choose 2 out of 8.
But there are only 28 different possible results, and 2 'ways'
to get each result.
