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I want to ask two questions:

(i) How many five-digit integers can one make from the digits $2, 2, 6, 0, 9, 4$?

(ii) How many digits in (i) satisfy that there is no $6$ followed by $2$?

My Approach

(i) I divided it into cases

  1. if we have $2$ in the first place so we have $1 \times 5 \times 4 \times 3 \times 2$ ways to choose the other 4 digits.

  2. if we have $2$ in the second place so we have $3 \times 1 \times 4 \times 3 \times 2$ ways to choose the 5 digits.

  3. if we have only one 2 so we have $3 \times 4 \times 4 \times 3 \times 2$ way to choose

and the result is the sum of the 3 cases.

(ii) I considered 62 as one digit, but couldn't solve it.

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2 Answers 2

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There seem to be some issues in your reasoning. The first question can be solved by considering three scenarios:

  1. The integer contains one $2$. There are four possibilities for the first digit (it cannot be $0$), after which the remaining four digits need to be assigned. The number of possibilities equals: $$4 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 96$$
  2. The integer contains two $2$s and no $0$. We choose the positions of the $2$s and then assign the positions of the remaining digits: $${5 \choose 2} \cdot 3 \cdot 2 \cdot 1 = 60$$
  3. The integer contains two $2$s and a $0$. We choose the position of the $0$, then choose the positions of the two $2$s, then choose the two remaining digits and finally assign their positions: $${4 \choose 1} \cdot {4 \choose 2} \cdot {3 \choose 2} \cdot 2 = 144$$

The total number of valid integers dus equals:

$$96 + 60 + 144 = 300$$

To answer the second question, we can consider all cases where the combination $62$ occurs. Using the same distinction as before:

  1. $62$ in front with $3 \cdot 2 \cdot 1 = 6$ possibilities or $62$ in second, third or fourth position, so assign $62$, then $0$, then the remaining digits for a total of $3 \cdot 2 \cdot 2 \cdot 1 = 12$ possibilities
  2. Consider $62$ as one of four digits, for a total of $4 \cdot 3 \cdot 2 \cdot 1 = 24$ possibilities
  3. $62$ in front with $3 \cdot {2 \choose 1} \cdot 2 \cdot 1 = 12$ possibilities or $62$ in second, third or fourth position, so assign $62$, then $0$, then the remaining digits for a total of $3 \cdot 2 \cdot {2 \choose 1} \cdot 2 \cdot 1 = 24$ possibilities

The number of possible integers thus equals:

$$300 - 6 - 12 - 24 - 12 - 24 = 222$$

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Assuming from your suggested solutions that digits can't be reused.

Then there are 5 choices for the first digit(discounting 0), 5 choices for the second digit, ...

Which gives us $5\cdot 5\cdot 4\cdot 3\cdot2=600$ possible combinations of digits. But since the 2 is used twice we must remove the repeated values from the total. Switching the position of the 2's produces the same number. (Whether switching inside the number or switching a 2 in the number with a left-over 2) Therefore the number of possible values is reduced by $\frac{1}{2}$.

$\frac{1}{2}\cdot 600=300$

For the 2nd part, take the complement of all the ways that have 2 following 6. so, $1 \cdot 2=2$ means that there are two ways to create 62 pair. Then 62 _ _ _ , _ 62 _ _, _ _ 62 _, and _ _ _ 62 gives $2\cdot 4\cdot 3\cdot 2=48$ and the remaining three options can't have lead zero so $3\cdot 2\cdot 3\cdot 2=36$. $48+36\cdot3=156$ 5-digit combinations with 6 followed by 2. Of these, swapping 2's reduces number by half and there are $\frac{156}{2}=78$ different numbers with a 62 in them. Which means there are $300-78=222$ different numbers that do not contain 6 followed immediately by 2.

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