Question

How many four-digit positive integers can be formed by using the digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two ?

Answer 1

Answer:

432

Step-by-step explanation:

We are given that a four digit number can be formed by using digits from 1 to 9.

Two digits equal to each other and other two digits are equal to each other .

We have to find number of four digit numbers can be formed using digits from 1 to 9.

Total digits =9

Number of ways to filled first place =9

Number of ways to filled second place =1  because two digits are same

Number of ways to fill the third place =8 because different from previous  two digits

Number of ways to fill the fourth place =1 because remaining two digits are same

Total number of ways to make four digit number =$9\times 1\times 8\times 1\times {\frac{4!}{2!2!}}$

Because  two digits out of 4  and  remaining two digits are same select out of 4

Total number of ways in which four digit number can be formed=$9\times 8\times 3\times 2=432$