0

# How many 7-digit numbers use exactly five different digits?

How many 7-digit numbers use exactly five different digits?

1
I think it would be : (9/10)*10C5*(5C2*(7!/(2!*2!))+5C1*(7!/3!))
9/10 takes care of leading 0 ( since all the digits have equal chance of being leading digit
10C5 selects first 5 different digits
5C2 selects 2 different digits from already used 5 digits to make the number look like (aabbcde)
7!/2!2! is the number of permutations for such numbers (aabbcde)
5C1 selects 1 digit from already used 5 digits to make the number look like (aaabcde)
7!/3! is the number of permutations for such numbers(aaabcde)
hence the final answer is (9/10)*10C5*(5C2*(7!/(2!*2!))+5C1*(7!/3!))
answered Oct 5, 2015 by Beginner (11 points)
0
(10*9*8*7*6*1*1) = 27216
27216 * ( 1+2+3+4+5 ) = 408240
&
30240 * 6 = 181440
so,
181440 + 408240 = 589680......
am i right ?????
answered Dec 9, 2014 by Beginner (1 point)
pls refer to my answer below and let me know if this is correct
commented Oct 5, 2015 by Partner (871 points)
Student Speaks

Jyotsna Mehta