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Question
Each of the digits 1, 1, 2, 3, 3 and 4 is written on a separate card. The six cards are then laid out in a row to form a 6-digit number. How many of these 6-digit numbers are divisible by 4?
Solution
In order to get the 6-digit number divisible by 4
The last two digits must be divisible by 4
∴ The last two digits should be 12 or 24 or 32
| 24 | ||||
| 1 | 2 | 3 | 4 | 5 |
Let the last box be filled with 24.
The remaining 4 boxes can be filled with the remaining digits
1, 1, 3, 3 in `(4!)/(2! xx 2!)` ways.
| 12 | ||||
| 1 | 2 | 3 | 4 | 5 |
Let the last box be filled with 12.
The remaining 4 boxes can be filled with the remaining digits.
1, 3, 3, 4 in `(4!)/(2!)` ways
| 32 | ||||
| 1 | 2 | 3 | 4 | 5 |
Let the last box be filled with 32.
The remaining 4 boxes can be filled with the remaining digits
The total number of 6 digit numbers which are divisible by 4 is
= `(4!)/(2! xx 2!) + (4!)/(2!) + (4!)/(2!)`
= `(1 xx 2 xx 3 xx 4)/(1 xx 2 xx 1 xx 2) + (1 xx 2 xx 3 xx 4)/(1 xx 2) + (1 xx 2 xx 3 xx 4)/(1 xx 2)`
= 6 + 12 + 12
= 30
∴ Required number of 6-digit numbers = 30
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| (a) How many numbers are formed? | (i) 840 |
| (b) How many number are exactly divisible by 2? | (i) 200 |
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