Question

four band members, harry, isabel, jose, and karen, are all boarding a bus to attend a concert. in each arrangement, the first initial of each person’s name represents that person’s position as he or she boards the bus: first, second, third, and fourth. if we know that karen boards the bus first, which shows the possible outcomes for this event? boxes with khij, khji, kjhi, kjih; khij; khij, khji, kjhi, kjih, kihj, kijh; khij, khji

Explanation:

Step1: Determine fixed position

Karen (K) boards first, so the first initial is always K. We need to find all permutations of the remaining three people: Harry (H), Isabel (I), Jose (J) for the second, third, and fourth positions.

Step2: Calculate permutations of 3 elements

The number of permutations of \( n \) distinct elements is \( n! \). Here \( n = 3 \), so \( 3! = 3\times2\times1 = 6 \)? Wait, no, wait. Wait, the remaining people are H, I, J. Wait, no, the four people are Harry (H), Isabel (I), Jose (J), Karen (K). If K is first, then we need to arrange H, I, J in the next three positions. The number of permutations of 3 elements is \( 3! = 6 \)? But wait, let's list them. The possible arrangements for the last three positions (second, third, fourth) with H, I, J:

• Second: H, Third: I, Fourth: J → KHIJ
• Second: H, Third: J, Fourth: I → KHJI
• Second: J, Third: H, Fourth: I → KJHI
• Second: J, Third: I, Fourth: H → KJIH
• Second: I, Third: H, Fourth: J → KIHJ
• Second: I, Third: J, Fourth: H → KIJH

Wait, but looking at the options, the third box has KHIJ, KHJI, KJHI, KJIH, KIHJ, KIJH. Wait, but wait, maybe I made a mistake. Wait, the four people: Harry (H), Isabel (I), Jose (J), Karen (K). So when K is first, the remaining three are H, I, J. The permutations of H, I, J are 6: H-I-J, H-J-I, I-H-J, I-J-H, J-H-I, J-I-H. Translating to initials:

• H-I-J → KHIJ
• H-J-I → KHJI
• I-H-J → KIHJ
• I-J-H → KIJH
• J-H-I → KJHI
• J-I-H → KJIH

So the set of possible outcomes should include all six permutations. Looking at the options, the third box has KHIJ, KHJI, KJHI, KJIH, KIHJ, KIJH, which matches the six permutations. Wait, but let's check the first box: KHIJ, KHJI, KJHI, KJIH – that's four, missing KIHJ and KIJH. The second box is just KHIJ – no. The fourth box is KHIJ, KHJI – two. So the correct one is the third box (the middle - right? Wait, the options are four boxes: first box: KHIJ, KHJI, KJHI, KJIH; second: KHIJ; third: KHIJ, KHJI, KJHI, KJIH, KIHJ, KIJH; fourth: KHIJ, KHJI. So the third box has all six permutations, which is correct because when K is fixed first, we arrange H, I, J in the next three positions, which has 3! = 6 permutations.

Answer:

The third box (containing KHIJ, KHJI, KJHI, KJIH, KIHJ, KIJH)