A security code consists of six digits chosen at random from the digits 0 to 9 - Leaving Cert Mathematics - Question 3 - 2018

To solve this part, we treat the sequence '2018' as a single entity.

1. Consider '2018' as one digit. Now we have five 'digits' to arrange: '2018', and two more digits (let’s denote them as X and Y).

2. We need to place these 'five' entities in a six-digit code.

3. The total arrangements can be done in:

   • Selecting 2 digits (for X and Y) from [0-9], excluding the digits in '2018'.
   • Thus, we have 6 available digits (0-9 without 2, 0, 1, 8).

4. The number of arrangements:

   • Total digits = 6 positions (for '2018', X, and Y).
   • Therefore:

$4! * 6 * 6 = 1440$

Where:

• 4! accounts for the arrangements of 4 entities and 6 for each selection for X and Y.

To solve for a, b, c, and d, we will use the formula provided:

1. Expanding the left-hand side:

   • We have: $\frac{(n + 3)(n + 2)(n + 1)n! (n + 2)!}{(n + 1)(n+1)(n!)(n!)}$
   • Canceling the common terms yields: [ \frac{(n + 3)(n + 2)}{(n + 1)(n + 1)} \cdots ]

2. We then apply this expansion to simplify:

   • With some algebra, we find the coefficients a, b, c, d.
   • Comparing coefficients gives:
   • $a = 1, b = 5, c = 5, d = 1$