6. (a) (i) In how many different ways can a committee of four people be selected from ten people? (ii) If one particular person must be on the committee, in how many different ways can the committee be selected? (b) Tickets for a raffle are placed in a box - Leaving Cert Mathematics - Question 6 - 2010

If one particular person must be included in the committee, we only need to select 3 more people from the remaining 9. Using the combination formula again:
$C(9, 3) = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84.$

To find the probability that the first ticket drawn is red, we use the formula for probability:
$P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{10}{25} = \frac{2}{5}.$

The probability that both tickets drawn are red is calculated as follows:

1. The probability that the first ticket drawn is red:
   $P(A) = \frac{10}{25}.$
2. The probability that the second ticket drawn is also red (after one red ticket has already been drawn):
   $P(B|A) = \frac{9}{24}.$
3. Thus, the required probability is:
   $P(A) \times P(B|A) = \frac{10}{25} \times \frac{9}{24} = \frac{15}{100} = \frac{3}{20}.$

To find this probability:

1. Probability first is red:
   $P(A) = \frac{10}{25}.$
2. Probability second is blue (after drawing a red ticket):
   $P(B|A) = \frac{15}{24}.$
3. Thus,
   $P(A) \times P(B|A) = \frac{10}{25} \times \frac{15}{24} = \frac{150}{600} = \frac{1}{4}.$

The possibilities are either a red followed by a blue or a blue followed by a red.

1. First red then blue:
   $\frac{10}{25} \times \frac{15}{24} = \frac{150}{600}.$
2. First blue then red:
   $\frac{15}{25} \times \frac{10}{24} = \frac{150}{600}.$
3. Thus, adding both:
   $\frac{150 + 150}{600} = \frac{300}{600} = \frac{1}{2}.$

The smallest possible four-digit code using the digits 3 to 9 is 3456.

To form a four-digit code using the digits 3 to 9 without repetition: there are 7 digits to choose from (3, 4, 5, 6, 7, 8, 9).
Thus:
$7 \times 6 \times 5 \times 4 = 840.$

To determine how many codes are greater than 6000, we consider digits available.
If the first digit is 6, 7, 8, or 9, we have:

• First digit: 4 options (6, 7, 8, 9).
• Remaining 3 digits: From the remaining 6 digits, we can choose 3.
  Thus, $4 \times 6 \times 5 \times 4 = 480.$

A four-digit number is divisible by 2 if its last digit is even (4, 6, 8).
Thus:

1. Last digit options: 3 (4, 6, 8).
2. First, second, and third digits can be chosen from the remaining 6 digits.
   Combination:
   For each case:
   $3 \times 6 \times 5 \times 4 = 360.$