Permutations and Combinations: Question 8

Worked solution

Part (a): No restriction other than distinct digits

Choosing and arranging $4$ of the $10$ digits, with none repeated, is a permutation:

$^{10}P_4 = 10\times9\times8\times7 = 5040.$

Part (b): Codes not starting with 0

Direct method: the first digit must come from $\{1,2,\ldots,9\}$ ($9$ choices, excluding $0$). Once the first digit is fixed, the remaining $3$ positions are filled from the $9$ digits still unused (which now includes $0$), arranged in order:

$9 \times\, ^9P_3 = 9\times(9\times8\times7) = 9\times504 = 4536.$

Check (complement): the codes that do start with $0$ have their first digit fixed as $0$, and the remaining $3$ positions filled from the other $9$ digits: $^9P_3 = 9\times8\times7=504$. Subtracting from the total in part (a):

$5040 - 504 = 4536.$

Both methods agree: $4536$.

Part (c): Valid codes (not starting with 0) that are even

Split into cases according to the last digit, since whether $0$ is “used up” by the last digit changes how many choices remain for the first digit.

Case 1: last digit is $0$. The last digit is fixed as $0$. The first digit must be nonzero, and since $0$ is already placed at the end, the first digit can be any of the remaining $9$ digits $\{1,\ldots,9\}$: $9$ choices. The middle two positions are then filled from the remaining $8$ digits, in order: $8\times7=56$. $9\times56 = 504.$

Case 2: last digit is $2$, $4$, $6$ or $8$ ($4$ choices of value). For each such value, the last digit is fixed (nonzero). The first digit must avoid both $0$ and the chosen last digit, so from the $9$ digits other than the last digit, $1$ (namely $0$) is excluded, leaving $8$ choices. The middle two positions are filled from the remaining $8$ digits (after removing the last digit and the first digit): $8\times7=56$. $8\times56 = 448 \text{ per last-digit value, so } 4\times448 = 1792.$

Total: $504 + 1792 = 2296.$

Check (via the odd case and part (b)‘s total): by the same casework, codes with an odd last digit ($1,3,5,7,9$, all nonzero) each give $8$ first-digit choices (excluding $0$ and the last digit) and $8\times7=56$ ways for the middle positions, so $5\times8\times56 = 2240$. Since every valid code from part (b) is either even or odd: $2296 + 2240 = 4536,$ which matches part (b) exactly, confirming $2296$.

Final answers

• (a) No restriction $= \boxed{5040}$
• (b) Not starting with $0$ $= \boxed{4536}$
• (c) Not starting with $0$ and even $= \boxed{2296}$