Question

arrangements with restrictions eg 3. from the digits 2, 3, 4, 5, 6 a) how many numbers greater than 4 000 can be formed? solution: 5 digits (any) = $_{5}p_{5}$ 4 digits (must start with digit ≥ 4) = $_{3}p_{1} \times _{4}p_{3}$ total = $_{5}p_{5} + _{3}p_{1} \times _{4}p_{3}$ b) how many 4 digit numbers would be even (ends with 2, 4 or 6) = $_{3}p_{1} \times _{4}p_{3}$ = $_{4}p_{3} \times _{3}p_{1}$

Explanation:

Part a) How many numbers greater than 40000 can be formed?

Step 1: Analyze the first digit requirement

To form a number greater than 40000 with 5 - digit numbers (using digits 2, 3, 4, 5, 6), the first digit must be 4, 5, or 6. So, there are 3 choices for the first digit.

Step 2: Arrange the remaining digits

After choosing the first digit, we need to arrange the remaining 4 digits. The number of permutations of \(n\) distinct objects taken \(r\) at a time is given by \(_{n}P_{r}=\frac{n!}{(n - r)!}\). Here, \(n = 4\) (remaining digits) and \(r=4\) (we need to arrange all 4 remaining digits). So, the number of ways to arrange the remaining 4 digits is \(_{4}P_{4}=\frac{4!}{(4 - 4)!}=\frac{4!}{0!}=4! = 24\).

Step 3: Calculate the total number of numbers

The total number of numbers greater than 40000 is the number of choices for the first digit multiplied by the number of arrangements of the remaining digits. So, total number \(=3\times_{4}P_{4}\)
\(=3\times24 = 72\)

Step 1: Analyze the last digit requirement

For a 4 - digit number to be even, the last digit must be 2, 4, or 6. So, there are 3 choices for the last digit.

Step 2: Arrange the first three digits

After choosing the last digit, we need to arrange the first 3 digits from the remaining 4 digits. The number of permutations of \(n = 4\) distinct objects taken \(r = 3\) at a time is \(_{4}P_{3}=\frac{4!}{(4 - 3)!}=\frac{4!}{1!}=4\times3\times2=24\).

Step 3: Calculate the total number of even 4 - digit numbers

The total number of even 4 - digit numbers is the number of choices for the last digit multiplied by the number of arrangements of the first three digits. So, total number \(=3\times_{4}P_{3}\)
\(=3\times24=72\)

Answer:

72

Part b) How many 4 - digit numbers would be even (ends with 2, 4 or 6)?