Find the number of distinguishable permutations of the letters m, i, s, s, i, s, s, i, p, p, i.
1 answer:
Solution:
we have been asked to find the number of distinguishable permutations of the letters m, i, s, s, i, s, s, i, p, p, i.
Here we can see
m appears 1 time.
i appears 4 times.
S appears 4 times.
p appears 2 times.
Total number of letters are 11.
we will divide the permutation of total number of letters by the permutation of the number of each kind of letters.
The number of distinguishable permutations
Hence the number of distinguishable permutations
You might be interested in
Answer:
Step-by-step explanation:
You're given a coordinate in the form of (x, y) and you're also given the vertex in the form of (h, k). We will use those in the vertex form of the equation
and solve for a. Filling in:
which simplifies a bit to
and a bit more to
. Add 3 to both sides to get
-8 = 4a so
a = -2
The equation, then, is
Answer:
The perimeter of other rectangles are 50, 28, 22 units
The area of all possible rectangles are same and equals to 24 unit²
Step-by-step explanation:
Given - Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units.
To find - What would be the perimeters of the other rectangles Orbet could make using only these 24 tiles? How do the areas of the rectangles compare?
Proof -
We know that
Are of Rectangle = Length × Breadth
And Perimeter of Rectangle = 2 (Length + Breadth )
Now,
Given that,
Orbet used 24 square tiles
And perimeter of the rectangle made = 20
So,
Possible Length of rectangle = 6
Possible breadth of Rectangle = 4
Or Vice-versa.
Total number of Rectangles possible = 4
Possibilities are - 1 × 24, 2 × 12, 3 × 8, 4 × 6
Case I :
Length of rectangle = 1
Breadth of rectangle = 24
∴ Perimeter of rectangle = 2(1 + 24) = 2(25) = 50 units
Area of rectangle = 1 × 24 = 24 unit²
Case II :
Length of rectangle = 2
Breadth of rectangle = 12
∴ Perimeter of rectangle = 2(2 + 12) = 2(14) = 28 units
Area of rectangle = 2 × 12 = 24 unit²
Case III :
Length of rectangle = 3
Breadth of rectangle = 8
∴ Perimeter of rectangle = 2(3 + 8) = 2(11) = 22 units
Area of rectangle = 3 × 8 = 24 unit²
Case IV :
Length of rectangle = 4
Breadth of rectangle = 6
∴ Perimeter of rectangle = 2(4 + 6) = 2(10) = 20 units
Area of rectangle = 4 × 6 = 24 unit²
∴ we get
The perimeter of other rectangles are 50, 28, 22 units
The area of all possible rectangles are same and equals to 24 unit²
