Answer:
1. 72
2. 216
3a. 2/17
3b. 1/425
Step-by-step explanation:
1.
Let's think about the first one a little bit. To simplify things, let's assume that Jimmy always sits down first, followed by Sean and then everyone else. If Jimmy sits at either end of the row, then Sean has 3 choices, followed by 3 choices for the next person, 2 for the next, and only 1 remaining for the last one. Therefore, we can divide this problem into two cases:
A) Jimmy sits at one of the sides
In this case, there are 2 choices for Jimmy, and then the amount of choices shown above, meaning that there are a total of 2*3*3*2*1=36 ways for them to be seated.
B) Jimmy sits in one of the center seats
In this case, there are two seats to either side of Jimmy that Sean will not sit in, leaving him two options. In this scenario, there are 2 options for Sean and 3 for Jimmy, which means there are a total of 3*2*3*2*1=36 ways for them to be sat.
Adding these two together, you get a total of 72 possible ways.
2.
Since all of the vowels cannot move around each other, we are essentially just finding all of the ways of arranging the consonants inside the word. Since the vowels must be stuck in the order AEIOU, then there are 6 positions for the letter Q, 6 for the letter T, and 6 for the letter N, making a total of 216 possible arrangements.
3a. The probability that the first card is red is 1/2, since half of the cards in the deck are red. Without replacement, the probability that the second card is red if the first one is will be 25/51, and the third card is 24/50. Multiplying these together, you get 25*24/50*51*2=2/17.
3b. It doesn't matter what the rank of the first card is, only whether the other two are the same rank. The probability of the second card being of the same rank as the first is 3/51, and for the third is 2/50, which when multiplied together gives 1/425. Hope this helps!
