Solution-
First 9 letters of alphabet are- A,B,C,D,E,F,G,H,I
Total number of ways of selection of 4 letters from 9 alphabets = 9C4
=9!÷((4!).(9-4)!) = 9!÷(4!×5!) = (9×8×7×6)÷(24) = 126
The number of ways of arranging these 4 numbers = 4! = 24
∴ Total number of possible permutations = 9C4×4! = 126×24 = 3024
∴ option number 2 is correct.
Step-by-step explanation:
= 5 • ( 4• x )
= 5 ( 4x )
= 20x

First you have to find the missing y
Answer:
option B
3
Step-by-step explanation:
Given in the question an expression

Whole numbers are positive numbers, including zero, without any decimal or fractional parts.
Possible range of domain will be 1 ≤ x ≤ 48
We know that perfect square between 1 and 48 are
1 , 4 , 16 , 25 , 36
1)
x = 3
√48/3 = √16 = 4
2)
x = 12
√48/12 = √4 = 2
3)
x = 48
√48/48 = √1 = 1
Answer:
The answer is C
Step-by-step explanation:
hope this helps