Step-by-step explanation:
a) 7!
If there are no restrictions, answer is 7! as it is the permutation of all animals.
b) 4! x 3!
As cats are 6 and Dogs are 5, thus 1st and last must be cats in order to have alternate arrangements. Therefore the only choices are the order of the cats among themselves and the order of the dogs among themselves. There
are 4! permutations of the cats and 3! permutations of the dogs,
so there are a total of 4! x 3! possible arrangements of the suites.
c) 3! x 5!
There are 3! possible arrangements of the dogs among themselves. Now, if we consider the dogs as one ”object” together, then we can think of arranging the 4 cats together with this 1 additional object. There are 5! such arrangements possible, so there are a total of 3! · 5! possible arrangements of the suites.
d) 2 x 4! x 3!
As required that all the cats must be together and all the dogs must be together, either the cats are all before the dogs or the dogs are all before the cats. There are two possible arrangements thus two times of both possibilities is the answer i.e. 2 x 4! x 3!
<span>(2n + 2)(2n – 2)
= (2n)^2 -2^2
= 4n^2 - 4
hope it helps</span>
3000000...........
10^6 =1000000 (3) = 3000000
This is an example of a distribution property. In a
distribution property, the number outside the parenthesis is distributed to all
the terms inside it by multiplication. So the answer here is:
<span>The 4 will be multiplied to each term inside the
parentheses.
</span>
Convection in the <u>Mantle</u> helps move pieces of the lithosphere around.
