kapil1 wrote:
Luca1111111111111 wrote:
Bunuel wrote:
How many unequal 5-digit numbers can be formed using each digit of the number 11235 only once?
(A) \(5!\)
(B) \(5P_3\)
(C) \(\frac{5C_5}{2!}\)
(D) \(\frac{5P_5}{2!*3!}\)
(E) \(\frac{5C_5}{2!*3!}\)
Can someone please explain:
I thought it is 4*3*2*1*4 = 96 because it is asked for an unequal 5-digit number meaning that the 2 cannot be the last digit?
5 digits can be arranged in 5! ways, but 2 digits are same so divide by 2! (as 2 digits can be rearranged within themselves)
5!/2! = 5P3
Sorry but this does not answer my question. I would like to know if we do not have to specifically change something in the calculation because we are looking for an unequal 5-digit number, meaning that we need to exclude the 2 as a possible last digit.
If we would change the question to ask for a 5-digit number with given values of 1,1,2,3,5 then I would assume that it is also 5P3 which cannot be as one time we look for an unequal number and another time for a number without restrictions