INTELLIGENT

There are 11 letters altogether, and 6 different types of letters.

2 x I, 2 x N, 2 x T, 2 x L, 2 x E ,1 x G

No restriction means the following cases are possible :

① All 4 letters different (eg. INTL)
② Two pairs of identical letters (eg. EELL)
③ 3 different types of letters, comprising
of 1 pair of identical letters, the other two letters are different (eg. NNTL)


①
Number of ways to choose 4 different types of letters from 6 choices
= 6C4
= 6! ÷ 4! ÷ (6 - 4)!
= 720 ÷ 24 ÷ 2
= 15

Number of ways to permutate the 4 letters = 4! = 24

Number possible 4-letter codewords for case ①
= 15 × 24
= 360

Or

Number of ways to choose 4 different letters from 6 choices and permutate them
= 6P4
= 6! ÷ (6-4)!
= 720 ÷ 2
= 360
This is actually part (i)'s answer


②
There are only 5 types of letters that have pairs (I,N,T,E,L)

Number of ways to select 2 types of letters from 5 choices for the pairs
= 5C2
= 5! ÷ 2! ÷ (5 - 2)!
= 120 ÷ 2 ÷ 6
= 10

Number of ways to permutate the two pairs of letters
= 4! ÷ 2! ÷ 2!
= 24 ÷ 2 ÷ 2
= 6

Number of possible 4-letter codewords for case ②
= 10 × 6
= 60


③
There is only one G so we can't do 6C3 immediately. Choose the pair first. Then choose the two different remaining letters.

Number of ways to choose 1 type of letter from 5 choices for the pair = 5

Number of ways to choose the remaining two different letters from the 5 remaining choices
= 5C2
= 10

Number of ways to permutate the pair and 2 different letters
= 4! ÷ 2!
= 24 ÷ 2
= 12

Number of possible 4-letter words for case ③
= 5 x 10 × 12
= 600

Total number of possible 4-letter codewords
= ①,②,③ combined
= 360 + 60 + 600
= 1020