Question -
Answer -
(i) We know that zerocannot be the first digit of the license plates. And the repetition of digitsis not allowed.
Let us assume five boxes,now the first box can be filled with one of the nine available digits, so thepossibility is┬а9C1
Similarly, the secondbox can be filled with one of the nine available digits, so the possibilityis┬а9C1
The third┬аbox canbe filled with one of the eight available digits, so the possibility is┬а8C1
The fourth┬аboxcan be filled with one of the seven available digits, so the possibilityis┬а7C1
The fifth┬аbox canbe filled with one of the six available digits, so the possibility is┬а6C1
Hence, the totalnumber of possible outcomes is┬а9C1┬а├Ч┬а9C1┬а├Ч┬а8C1┬а├Ч┬а7C1┬а├Ч┬а6C1┬а=9 ├Ч 9 ├Ч 8 ├Ч 7 ├Ч 6 = 27,216
(ii)┬аWe know thatzero cannot be the first digit of the license plates. And the repetition ofdigits is allowed.
Let us assume fiveboxes, now the first box can be filled with one of the nine available digits,so the possibility is┬а9C1
Similarly, the secondbox can be filled with one of the ten available digits, so the possibilityis┬а10C1
The third┬аbox canbe filled with one of the ten available digits, so the possibility is┬а10C1
The fourth┬аboxcan be filled with one of the ten available digits, so the possibility is┬а10C1
The fifth┬аbox canbe filled with one of the ten available digits, so the possibility is┬а10C1
Hence, the totalnumber of possible outcomes is┬а┬а9C1┬а├Ч┬а10C1┬а├Ч┬а10C1┬а├Ч┬а10C1┬а├Ч┬а10C1┬а=9 ├Ч 10 ├Ч 10 ├Ч 10 ├Ч 10 = 90,000