The word TOADSTOOL consists of the letters: T, O, A, D, S, T, O, O, L.

We need to select 5 letters including at least 2 Os and at least 1 T.

Consider the following cases:

1. Case 1: 2 Os and 1 T (OOT__). We need 2 more letters from A, D, S, L.

   The number of ways to choose 2 more letters from 4 is given by: \(\binom{4}{2} = 6\).

2. Case 2: 2 Os and 2 Ts (OOTT_). We need 1 more letter from A, D, S, L.

   The number of ways to choose 1 more letter from 4 is given by: \(\binom{4}{1} = 4\).

3. Case 3: 3 Os and 1 T (OOOT_). We need 1 more letter from A, D, S, L.

   The number of ways to choose 1 more letter from 4 is given by: \(\binom{4}{1} = 4\).

4. Case 4: 3 Os and 2 Ts (OOOTT). No more letters needed.

   There is only 1 way to choose this combination.

Adding all the cases together gives: \(6 + 4 + 4 + 1 = 15\).

The word STEEPLECHASE consists of the letters: S, T, E, E, P, L, E, C, H, A, S, E.

We need to select four letters including exactly one S. There are two S's in the word, so we choose one S and three other letters from the remaining 10 letters.

First, count the distinct letters after choosing one S: T, E, P, L, C, H, A, E, E, E.

We can choose the remaining three letters in the following ways:

1. Choose 3 E's: inom{6}{1} = 6
2. Choose 2 E's and 1 other letter: inom{6}{2} = 15
3. Choose 1 E and 2 other letters: inom{6}{3} = 20

Adding these scenarios gives the total number of selections:

\(6 + 15 + 20 = 41\)

Therefore, the total number of different selections is 42.

(iv) To find the number of selections with exactly one M and one A, we choose 1 M, 1 A, and 1 other letter from the remaining letters.

The word CAMERAMAN has the letters C, A, M, E, R, A, M, A, N.

There are 3 A's and 2 M's. We choose 1 M and 1 A, and then choose 1 more letter from the remaining 6 letters (C, E, R, N, A, M).

The number of ways to choose 1 letter from these 6 is given by:

\(\binom{4}{1} = 4\)

Thus, the number of different selections is 4.

(v) To find the number of selections with at least one M, we consider different cases:

1. Case 1: 1 M and 2 other letters (not both A's). Choose 1 M and 2 from the remaining 7 letters (C, A, E, R, A, N).

\(\binom{4}{2} = 6\)

2. Case 2: 2 M's and 1 other letter. Choose 1 from the remaining 7 letters (C, A, E, R, A, N).

\(\binom{4}{1} = 4\)

3. Case 3: 1 M, 1 A, and 1 other letter. Already calculated as 4.

\(Total number of selections = 6 + 4 + 4 = 16.\)

The word MISSISSIPPI consists of the following letters: M, I, S, S, I, S, S, I, P, P, I.

\(Count of each letter: M = 1, I = 4, S = 4, P = 2.\)

Total number of ways to choose 2 letters from 11: \(\binom{11}{2} = 55\).

Ways to choose 2 identical letters:

• 2 S's: \(\binom{4}{2} = 6\)
• 2 I's: \(\binom{4}{2} = 6\)
• 2 P's: \(\binom{2}{2} = 1\)

Total ways to choose 2 identical letters: \(6 + 6 + 1 = 13\).

Probability that the two letters are the same: \(\frac{13}{55}\).

(iii) To find the number of selections with exactly 2 Es and 2 Ns, we consider the letters in SEVENTEEN: S, E, V, E, N, T, E, E, N. We need to select 2 Es and 2 Ns, leaving 1 more letter to choose from S, V, or T. This can be done in 3 ways: EENN + S, EENN + V, EENN + T. Thus, there are 3 possible selections.

(iv) To find the number of selections with at least 2 Es, we consider the following cases:

1. 2 Es: Choose 2 Es and 3 other letters from S, V, N, T, N. This can be done in 7 ways: EENN + S, EENN + V, EENN + T, EENT + S, EENT + V, EENT + N, EENS + V.
2. 3 Es: Choose 3 Es and 2 other letters from S, V, N, T, N. This can be done in 7 ways: EEE + SV, EEE + SN, EEE + ST, EEE + VN, EEE + VT, EEE + NT, EEE + NN.
3. 4 Es: Choose 4 Es and 1 other letter from S, V, N, T. This can be done in 4 ways: EEEE + S, EEEE + V, EEEE + N, EEEE + T.

\(Adding these, the total number of ways is 7 + 7 + 4 = 18.\)