How many 4-digit numbers can be formed by using the digits 0-9, so that the numbers contains exactly 3 distinct digits?  Choose the correct choice from the following:	A) 1944	B) 3240	C) 3850	D) 3888	E) 4216
Case I:   The repeated digit is the unit's digit.   So, the 1st, 2nd and 3rd digits can be selected in 9 x 9 x 8 ways, respectively.   Now the 4th digit (unit's digit) can be either equal to the 1st, 2nd or 3rd digit.   Thus, in all we have:   9x9x8x3   Case II:   The repeated digit is the ten's digit.   So, the 1st, 2nd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.   Now the 3rd digit (ten's digit) can be either equal to the 1st or 2nd digit.   Thus, in all we have:   9x9x2x8   Case III:   The repeated digit is the hundred's digit.   So, the 1st, 3rd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.   Now the 2nd digit (hundred's digit) is equal to the 1st digit.   Thus, in all we have:   9x1x9x8   In totality, we have 9x9x8(3+2+1) = 9x9x8x6 = 3888   ANS:3888
<result>D</result>
How many 4-digit numbers can be formed by using the digits 0-9, so that the numbers contains exactly 3 distinct digits?