How many 6 digits numbers contain exactly 4 different digits?
Answers
Step-by-step explanation:
The no. of permutation of 4 digits taken at a random is
9*9*8*7 now we select 2 more digits from the 4 digit already selected, this could make the no. in format of abcd[ab] or abcd[aa].
There are two permutations for them 4*3 or 4*1.
=(9*9*8*7)[4*3+4*1] ,
BUT the 1st four digits have a permutation of their own and the last two digits have a permutation of their own ,they are like 2 separate units, we want a permutation of 6 digits as a whole and not 4 digits and 2 digits. so we remove the unwanted permutation.
=(9*9*8*7/4!)[(4*3)/2!) + (4*1)/1!], 1! because in 2nd case the two no. are the same
now we do permutations of 6 digits to both cases, in first case you have 2 identical pairs and in second case you have 1 identical group of 3
=(9*9*8*7/4!)[4*3/2! * (6!/2!2!) + 4*1/1! * (6!/3!)]
= 294840