Answer:
a.4845ways.
b. 14535ways.
c. 3990ways
d. 1140ways
Step-by-step explanation:
Given data:
No of flavors available to customers = 20.
Solution:
This is permutation and combinations problem,
(a) how many ways can the customers choose 4 different ice creams if they are all of different flavors.
20C4
= n!/(n-k)!)k!
= 20!/(20-4)!)4!
= 20!/(16)!)4!
= 4845ways.
b) are not necessarily of different flavors
Let’s say any two same flavors can be chosen.
20C4 * 3!/2!
= 4845 * 3
= 14535ways.
c) contain only 2 or 3 flavors.
= 20C3 * 3!/2!
= 1140 * 3
= 3420
20C2 * 3
= 190 * 3
= 570.
No of 2 or 3 different flavors
= 3420 + 570
= 3990ways.
d) contain 3 different flavors.
20C3
= n!/(n-k)!)k!
= 20!/(20-3)!)3!
= 1140ways.
Give us info on the deck if you want an answer
Answer:
Option C. f(n) = 16(3/2)⁽ⁿ¯¹⁾
Step-by-step explanation:
To know which option is correct, do the following:
For Option A
f(n) = 3/2(n – 1) + 16
n = 1
f(n) = 3/2(1 – 1) + 16
f(n) = 3/2(0) + 16
f(n) = 16
n = 2
f(n) = 3/2(n – 1) + 16
f(n) = 3/2(2 – 1) + 16
f(n) = 3/2(1) + 16
f(n) = 3/2 + 16
f(n) = 1.5 + 16
f(n) = 17.5
For Option B
f(n) = 3/2(16)⁽ⁿ¯¹⁾
n = 1
f(n) = 3/2(16)⁽¹¯¹⁾
f(n) = 3/2(16)⁰
f(n) = 3/2 × 1
f(n) = 1
For Option C
f(n) = 16(3/2)⁽ⁿ¯¹⁾
n = 1
f(n) = 16(3/2)⁽¹¯¹⁾
f(n) = 16(3/2)⁰
f(n) = 16 × 1
f(n) = 16
n = 2
f(n) = 16(3/2)⁽ⁿ¯¹⁾
f(n) = 16(3/2)⁽²¯¹⁾
f(n) = 16(3/2)¹
f(n) = 16(3/2)
f(n) = 8 × 3
f(n) = 24
n = 3
f(n) = 16(3/2)⁽ⁿ¯¹⁾
f(n) = 16(3/2)⁽³¯¹⁾
f(n) = 16(3/2)²
f(n) = 16(9/4)
f(n) = 4 × 9
f(n) = 36
For Option D
f(n) = 8n + 8
n = 1
f(n) = 8(1) + 8
f(n) = 8 + 8
f(n) = 16
n = 2
f(n) = 8n + 8
f(n) = 8(2) + 8
f(n) = 16 + 8
f(n) = 24
n = 3
f(n) = 8n + 8
f(n) = 8(3) + 8
f(n) = 24 + 8
f(n) = 32
From the above illustration, only option C describes the sequence.
Answer:
d
Step-by-step explanation:
Under a reflection in the y-axis
a point (x, y) → (- x, y )
<em>they </em><em>both </em><em>would </em><em>have </em><em>2</em><em>4</em>
<em> </em><em>they </em><em>both </em><em>would </em><em>have </em><em>2</em><em> </em><em>successful</em><em> </em><em>shots.</em><em> </em><em>they </em><em>would </em><em>mean </em><em>that </em><em>the </em><em>older </em><em>brother</em><em> </em><em>added </em><em>4</em><em> </em><em>point </em><em>to </em><em>his </em><em>score </em><em>and </em><em>the </em><em>younger </em><em>brother</em><em> </em><em>would </em><em>have </em><em>added </em><em>6</em><em> </em><em>to </em><em>his </em><em>score</em>
