combinatorics - letter arrangements (1 Viewer)

[waltssillyhat]

Using the letters P,P,Q,R,S,T,T, how many codes can be formed if they contain exactly four of the seven letters?

i understand how to do this question if the codes contain all seven letters, but how would i approach this if they only contain four? the answer in the book is 5P4 + 2 x 4C2 x 4!/2! +4!/2!2! but i can't make sense of it. any help is appreciated <33

 

[Legendary16]

Case 1: all letters are different. Here we have 5 different letters (P,Q,R,S,T) and so we pick 5 letters to arrange in 4 spaces. Hence we have 5P4.
Case 2: One letter is repeated. Since P and T are repeated, this case will be doubled for either letter. For P, since 2 letters are are already chosen, 4C2 would select 2 letters from (Q,R,S,T) without arrangement. Now these letters can be arranged in 4! ways, though P is repeated here. Hence it is divided by 2!. Summatively, we thus have 2 (for P and T) x 4C2 x 4!/2! = 4! x 4C2
Case 3: Two letters are repeated (P,P,T,T). Now, we have 4 letters to arrange but since 2 are repeated, we have 4!/2!2!

 

[waltssillyhat]

Case 1: all letters are different. Here we have 5 different letters (P,Q,R,S,T) and so we pick 5 letters to arrange in 4 spaces. Hence we have 5P4.
Case 2: One letter is repeated. Since P and T are repeated, this case will be doubled for either letter. For P, since 2 letters are are already chosen, 4C2 would select 2 letters from (Q,R,S,T) without arrangement. Now these letters can be arranged in 4! ways, though P is repeated here. Hence it is divided by 2!. Summatively, we thus have 2 (for P and T) x 4C2 x 4!/2! = 4! x 4C2
Case 3: Two letters are repeated (P,P,T,T). Now, we have 4 letters to arrange but since 2 are repeated, we have 4!/2!2!

this makes sense thank you so much!!