q40

S

Saumyojit

Senior Member
Joined
Jan 21, 2020
Messages
1,032
How many four digit numbers are there whose decimal notation contains not more than two distinct digits?

Case 1(only one Digit) : _ _ _ _ : 9 ways

Case 2(two digits without 0) : __ _ _ _ : (9c1 * 4 )* 8c1 =288ways eg: 1222,2122,2212 , 2221, 2111,1211 etc

Case 3 : 0 repeated and any other digit : 9 ways eg : 1000 , 2000, ...9000

Case 4: 0 single and other digit repeated : 9c1 * 3 =27 eg: 2202 ,2022,2220 etc

What am i missing?
answer is 576
 
Case 5 : 2 of one type and 2 of another except 0 : 9c1 * 8c1 * 6= 432 eg: 1122, 1221 etc

Case 6: 2 of one type and two zero : 9c1 * (3*2)/2 = 27 ways eg: 1100 , 1010 etc
 
Saumyojit said:
Case 5 : 2 of one type and 2 of another except 0 : 9c1 * 8c1 * 6= 432 eg: 1122, 1221 etc
Click to expand...
Change 9c1 * 8c1 (which is 9P2) to 9C2, and you'll get the right total. (I just looked for a large number in your work that could be reduced.)

But is that correct? Your 6 here is 4C2, choosing the two places for one digit to go, so that the choice of the two digits shouldn't distinguish them; given a pair like {1,2}, without regard to order, the 6 will include both 1122 and 2211, for example. These shouldn't be counted twice, once for (1,2) and again for (2,1).

Frankly, I'm having trouble convincing myself! Therefore, I look for another way to get the same answer: We can first choose the first digit (9 ways), then choose the other place to put it (3 ways), then choose the digit to go in the other 2 places (8 ways), total 9*3*8 = 216. Now I'm convinced: it isn't 432.

If I were you, I'd now try to simplify the work, by arranging your cases more neatly, at the least. As it is, it's not easy to follow, which makes it hard to be sure it's right. Fewer cases, more clearly listed, would help.
 
You must log in or register to reply here.
Top