Answer:
So i say C
Step-by-step explanation:
240 = area
Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
Answer: 12 friends.
Step-by-step explanation:
the data we have is:
Mei Su had 80 coins.
She gave the coins to her friends, in such a way that every friend got a different number of coins, then we have that:
The maximum number of friends that could have coins is when:
friend 1 got 1 coin
friend 2 got 2 coins
friend 3 got 3 coins
friend N got N coins
in such a way that:
(1 + 2 + 3 + ... + N) ≥ 79
I use 79 because "she gave most of the coins", not all.
We want to find the maximum possible N.
Then let's calculate:
1 + 2 + 3 + 4 + 5 = 15
15 + 6 + 7 + 8 + 9 + 10 = 55
now we are close, lets add by one number:
55 + 11 = 66
66 + 12 = 78
now, we can not add more because we will have a number larger than 80.
Then we have N = 12
This means that the maximum number of friends is 12.
You solve by with by altitude! Hope it helps
