How many numbers can be formed from the digits 0, 1, 3, 4, 5, 7, and 9 taken 5 at a time, if every number must contain the digits 0,4,5?
For me, I utilized the permutation method, as 5 taken digits can be permuted between them upon choices.
But, every number must contain 0,4,5 so 3 places are preoccupied. 0 can be in 5 places (as the number of the digits of the number to be formed has not been defined in the question), 4 can be in 4 places, 5 can be in 3 places. The remaining two places can be filled by 1, 3, 7, and 9, so $^4P_2$.
Thus, my answer would be 5 * 4 * 3 * 4 * 3. Can somebody pls help me with this problem?
Best Answer
As we know, $0$, $4$ & $5$ have to be in the number formed, so the other two digits have to be chosen from the remaining four options $1$, $3$, $7$ or $9$. Also, we will have to arrange these five digits to get all the possible permutations for this question.
Hence, the answer expression shall be:
$^4C_2 * 5!$
= $6*120$
= $720$
Kindly note that
did not imply that the number has to be a 5 digit number, ie. $0$ can be placed at the first position too! Else, $5!$ should be replaced by $4*4!$