Answer:
There are 15 letters, but if the two A's must always be together, that's the same as if they're just one letter, so our "base count" is 14! ; note that this way of counting means that we also don't need to worry about compensating for "double counting" identical permutations due to transposition of those A's, because we don't "count" both transpositions. However, that counting does "double count" equivalent permutations due to having two O's, two N's, and two T's, so we do need to compensate for that. Therefore the final answer is 14!/(23)=10,897,286,400
Answer:
x = 9
Step-by-step explanation:
The product of the external part and the entire part of one secant is equal to the product of the external part and the entire part of the other secant, that is
5(x - 6 + 5) = 4(x - 3 + 4)
5(x - 1) = 4(x + 1) ← distribute parenthesis on both sides
5x - 5 = 4x + 4 ( subtract 4x from both sides )
x - 5 = 4 ( add 5 to both sides )
x = 9
Answer:
50
Step-by-step explanation:
1) 15
2)6
3)9
Hope this helps!
I think with the kinds of mixture problems, it's really helpful to make a table.
The equations you actually need are
and
. Since I was so close, I went ahead and just solved it anyway. You can see my setup on the attached image.
