# how many five-digit integers can one make from the digits 2,2,6,0,9,4?

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.

• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. Mar 1, 2020 at 23:11

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:

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

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.