Yeah, this exercise is a bit tricky but good training for nested for-loops.  

Best,

function solve(input) {
    let lastSector = input.shift();
    let initialRowCounter = Number(input.shift());
    let seatCounter = Number(input.shift());
    // let firstSectorNum = Number(input.shift());
    //let numOddSeats = Number(input.shift());
    let firstCharNum = 'a'.charCodeAt(0);
    let totalSeats = 0;

    let firstChar = 'A';
    let str = input;
    for (let i = firstChar.charCodeAt(0); i <= lastSector.charCodeAt(0); i++) {
        for (let row = 1; row <= initialRowCounter; row++) {
            if (row % 2 !== 0) {
                for (let j = firstCharNum; j < firstCharNum + seatCounter; j++) {
                    let currentLetter = String.fromCharCode(i);
                    let currentSmallLetter = String.fromCharCode(j);
                    console.log(`${currentLetter}${row}${currentSmallLetter}`);
                    totalSeats++;
                }
            } else {
                for (let j = firstCharNum; j < firstCharNum + seatCounter + 2; j++) {
                    let currentLetter = String.fromCharCode(i);
                    let currentSmallLetter = String.fromCharCode(j);
                    console.log(`${currentLetter}${row}${currentSmallLetter}`);
                    totalSeats++;
                }
            }
        }

        initialRowCounter++;

        // for (let j = 1; j <= firstSectorNum; j++) {

        //     str = (String.fromCharCode(i) + " " + j);
        //     if (j % 2 !== 0) {
        //         console.log(str);
        //         console.log(str);
        //     } else if (j % 2 === 0) {
        //         console.log(str);
        //         console.log(str);
        //         console.log(str);
        //         console.log(str);

        //     }
        // }
    }
    console.log(totalSeats);

}


solve(['B', '3', '2']);
solve(['C', 4, 2]);