How many numbers can be considered of five figures (from 10000 to 99999) if we require that exactly four different figures appear in nondecreasing order (example: 23779)?
The solution is 4*C(9,4). I understand the *4 since we are gonna choose the number which will be equal to one of the other 4, but why C(9,4)?
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$\begingroup$Let the five digit number is $$X_1X_2X_3X_4X_5$$ where condition is that of five digits 4 are distinct and there exist inequality $X_1\le X_2 \le X_3 \le X_4 \le X_5$$$ $$Consider along with nine digits 1,2,4,....,9 assume four more distinct objects $X_{(1,2)},X_{(2,3)},X_{(3,4)},X_{(4,5)}$ where $X_{(i,i+1)}$ indicate digit $X_i \,and \,X_{i+1}$ are equal.$$ $$Number of ways are from 9 digit select any 4 and from different $X_{(i,i+1)}$ select 1 we get required number $$={9 \choose 4}{4 \choose 1}$$$$=504$$
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