In this question, a group of number/symbol is coded using letter codes as per the codes given below and the conditions which follow. The correct combination of codes following the conditions is your answer. If none of the conditions follow, then codes for
the respective Number/Symbol to be followed directly as given in the table.

NUMBER/SYMBOL     7$86&#3+5429@*
CODE                              AERTHUKLPZCVNQ
Conditions
(i) If the first element is a symbol and the last a number, the codes for these two (the first and the last elements) are to be interchanged.
(ii) If the first element is an odd number and the last an even number, the first and last elements are to be coded as ©
(iii) If both second and third elements are perfect squares, the third element is to be coded as the code for the second element.
What will be the code for the following group?
7 4 9 @ $

To find the correct code combination for the group 7 4 9 @ $, let us first analyze the standard character-to-code mapping provided in the table:

NUMBER/SYMBOL: 7 | $ | 8 | 6 | & | # | 3 | + | 5 | 4 | 2 | 9 | @ | *
CODE: A | E | R | T | H | U | K | L | P | Z | C | V | N | Q

From this mapping, the individual codes for each element in 7 4 9 @ $ are:

• 7 → A
• 4 → Z
• 9 → V
• @ → N
• $ → E

Now, let us evaluate the given conditions step-by-step to see if any apply to 7 4 9 @ $:

Condition (i): "If the first element is a symbol and the last a number, the codes for these two are to be interchanged."
Here, the first element is 7 (a number) and the last element is $ (a symbol). This condition does not apply.

Condition (ii): "If the first element is an odd number and the last an even number, the first and last elements are to be coded as ©."
The first element is 7 (odd), but the last element is $ (symbol, not an even number). This condition does not apply.

Condition (iii): "If both second and third elements are perfect squares, the third element is to be coded as the code for the second element."
The second element is 4 (which is 2², a perfect square) and the third element is 9 (which is 3², a perfect square).
Since both the 2nd and 3rd elements are perfect squares, this condition applies!
Therefore, the third element (9) must take the code of the second element (4), which is Z.

Applying Condition (iii) to our initial code sequence:

• 1st element: 7 → A
• 2nd element: 4 → Z
• 3rd element: 9 → Z (changed from V to match the 2nd element's code)
• 4th element: @ → N
• 5th element: $ → E

Combining all the coded elements gives the final sequence: A Z Z N E.

The correct option is A Z Z N E.