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Q.
How many 5 digit numbers are there which contains not more than two different digits ?
Permutations and Combinations
Solution:
Excluding the digit ' 0 ', two digits out of the remaining 9 can be selected in ${ }^9 C _2$ ways. e.g. $12,23,34$ etc.
Now all the 5 digit numbers which can be made which do not contain all digits identical $={ }^9 C _2\left(2^5-2\right)$
2 has been deducted from $2^5$ to exclude numbers formed by all five alike digits otherwise their will be repeatition when we take 1 and 2 then 11111 will appear & when we select 1 and 3 then again 11111 will appear.
( like 1111111$)$
But we have 9 such numbers containing all alike digits. Hence total 5 digit numbers none of them containing the digit ' 0 ' having not more than two alike digits,
$={ }^9 C_2\left(2^5-2\right)+9=1080+9=1089$
Now with ' 0 ' always included, we have select the remaining 1 digit in ${ }^9 C _1$ ways e.g. 01,02 etc.
[Note that 1st place can be filled only in one way and the remaining digits in two ways]
$(1)+(2) 1080+144=1224$ (Ans.)
(Note that 9 has been subtracted as all five digits identical has again been counted)
