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 :)
Answer:
The answer to your question is 113.04 in²
Step-by-step explanation:
To solve this problem substitute the values in the formula and simplify.
Data
Area = ?
radius = 6 in
π = 3.14
Formula
Area = π x r²
Substitution
Area = (3.14) x (6)²
Simplification
Area = 3.14 x 36
Result
Area = 113.04 in²
Answer:
1,5
Step-by-step explanation:
Answer:
(a)
(b)
(c)43 times
Step-by-step explanation:
<u>Part A</u>
The paper's thickness = 0.05mm
When the paper is folded, its width doubles (increases by 100%).
The thickness of the paper grows exponentially and can be modeled by the function:
<u>Part B</u>
<u />
<u />
<u />
<u>Part C</u>
If the thickness of the paper, g(n)=384,472,300,000 mm
Then:
You must fold the paper 43 times to make the folded paper have a thickness that is the same as the distance from the earth to the moon.
