Question

miriam is visiting five friends in her neighborhood, louise, mike, nicole, oscar, and pascal. in each arrangement, the first initial of each person’s name represents that person’s position as he or she is visited by miriam. which shows all the outcomes for this event if she visits with nicole first and louise last? nlmop, nmlop, nolmp; nlmop, nlomp, nmlop, nmolp, nomnp, nonmp; nmopl, nmpol, nopml, nompl, npmol, npoml; nmopl, nompl, npmol

Explanation:

Step1: Identify fixed positions

Nicole (N) is first, Louise (L) is last. So each outcome should start with N and end with L.

Step2: Analyze remaining friends

Remaining friends: Mike (M), Oscar (O), Pascal (P). We need to arrange M, O, P in the middle three positions (2nd, 3rd, 4th).
The number of permutations of 3 elements is \(3! = 6\)? Wait, no, wait: Wait, the friends are Mike (M), Oscar (O), Pascal (P) – three people. Wait, but let's check the options. Wait, the correct outcomes should start with N and end with L, with M, O, P in positions 2,3,4. So the structure is N _ _ _ L. The middle three positions are for M, O, P. The permutations of M, O, P are:

1. M, O, P → N M O P L → NMOPL
2. M, P, O → N M P O L → NMPOL
3. O, M, P → N O M P L → N O M P L? Wait, no, wait: Wait, the initials: Mike is M, Oscar is O, Pascal is P. Wait, no, wait: Wait, the friends are Mike (M), Oscar (O), Pascal (P). Wait, but let's check the options. The third option has NMOPL, NMPOL, NOPML, NOMPL, NPMOL, NPOML. Wait, let's check each:

• NMOPL: N (1st), M (2nd), O (3rd), P (4th), L (5th) – correct.
• NMPOL: N, M, P, O, L – correct.
• NOPML: N, O, P, M, L – correct.
• NOMPL: N, O, M, P, L – correct.
• NPMOL: N, P, M, O, L – correct.
• NPOML: N, P, O, M, L – correct.

Now check other options:
First option: NLMOP (ends with P, not L) – invalid.
Second option: NLMOP (ends with P), NLOMP (ends with P) – invalid.
Fourth option: Only three outcomes, but we have 3! = 6 permutations? Wait, no, wait: Wait, the friends are Mike (M), Oscar (O), Pascal (P) – three people, so 3! = 6 permutations. Wait, but let's count:
Permutations of M, O, P:

1. M, O, P → NMOPL
2. M, P, O → NMPOL
3. O, M, P → N O M P L? Wait, no, N O M P L is NOMPL? Wait, N O M P L: N (1), O (2), M (3), P (4), L (5) → NOMPL.
4. O, P, M → N O P M L → NOPML.
5. P, M, O → N P M O L → NPMOL.
6. P, O, M → N P O M L → NPOML.

Yes, that's 6 permutations, which matches the third option. Wait, but wait, the third option has six outcomes: NMOPL, NMPOL, NOPML, NOMPL, NPMOL, NPOML. Let's check each:

• NMOPL: N-M-O-P-L – correct.
• NMPOL: N-M-P-O-L – correct.
• NOPML: N-O-P-M-L – correct.
• NOMPL: N-O-M-P-L – correct.
• NPMOL: N-P-M-O-L – correct.
• NPOML: N-P-O-M-L – correct.

Other options:
First option: NLMOP (ends with P), NML0P (typo? NML0P? No, NML0P – ends with P), NOLMP (ends with P) – all end with P, not L – invalid.
Second option: NLMOP (ends with P), NLOMP (ends with P), NML0P (ends with P), NMOLP (ends with P), NOMNP (ends with P), NONMP (ends with P) – all end with P, not L – invalid.
Fourth option: NMOPL, NOMPL, NPMOL – only three, but we need six? Wait, no, wait: Wait, the friends are Mike (M), Oscar (O), Pascal (P) – three people, so 3! = 6 permutations. So the third option has six outcomes, all starting with N and ending with L, with M, O, P in the middle. So that's correct.

Answer:

The third option (NMOPL, NMPOL, NOPML, NOMPL, NPMOL, NPOML)