Find the number of different ways in which the 10 letters in the word AMALGAMATE can be arranged so that...

Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

1. Method 1 M _ _ _ _ _ _ M (arranging the remaining letters and inserting the As): [M1] 4! × m, m∈Z,m≥1. [M1] n × 5C1 or n × 5C4 or n × 5, allow n × 5P1 or n × 5P4/4! , n∈Z,n>1. e.g. 4! × 5C1× k (oe), k∈Z, k≥ 1 scores M1M1. [A1] 120.
2. Method 2 Total number of arrangements with Ms at end – arrangements with Ms at end and at least 2 As together: [M1] 8!/4! -xr-5!×m, m∈Z,m ≥ 3,r = 1,2 . [M1] k-5!×13×r, k∈Z,k>1560r=1,2 . [A1] 120.
3. Method 1 M _ _ _ M _ _ _ _ _ : [B1] 8!/4! ×b,b∈Z,b≥1. [M1] c!/d ×6, c = 8, 9, 10 d = 1, 2!, 4! or 2! × 4!. [A1] 10080.
4. Method 2 M^^^M letters between Ms arranged and the treated as a single item: [B1] 6!/4! ×b,b∈Z,b≥1. [M1] c!/d x8P3, c = 6,7,8 d = 1, 1, 2!, 4! or 2!×4!. [A1] 10080.
5. Method 1: [B1] One identified outcome correct (excluding MMAAA or MAAAA), accept un-simplified. [M1] Five correct scenarios added (values do not need to be correct). [A1] 16.
6. Method 2: 2 Ms cannot be present in the scenarios: [B1] One identified outcome correct (excluding MAAAA), accept un-simplified. [M1] Three correct scenarios added (values do not need to be correct). [A1] 16.