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 9C1
Similarly, the secondbox can be filled with one of the nine available digits, so the possibilityis 9C1
The third box canbe filled with one of the eight available digits, so the possibility is 8C1
The fourth boxcan be filled with one of the seven available digits, so the possibilityis 7C1
The fifth box canbe filled with one of the six available digits, so the possibility is 6C1
Hence, the totalnumber of possible outcomes is 9C1 × 9C1 × 8C1 × 7C1 × 6C1 =9 × 9 × 8 × 7 × 6 = 27,216
(ii) We 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 9C1
Similarly, the secondbox can be filled with one of the ten available digits, so the possibilityis 10C1
The third box canbe filled with one of the ten available digits, so the possibility is 10C1
The fourth boxcan be filled with one of the ten available digits, so the possibility is 10C1
The fifth box canbe filled with one of the ten available digits, so the possibility is 10C1
Hence, the totalnumber of possible outcomes is 9C1 × 10C1 × 10C1 × 10C1 × 10C1 =9 × 10 × 10 × 10 × 10 = 90,000