WINNIPEG PERMUTATION

Question

In how many ways can the letters of the word WINNIPEG be arranged if the letters W, P, and G must remain in the same order they appear in the word?

Answer 1

Since W, P, and G must remain in the same order but not location, we can count the number of unique arrangements of the blanks in sequences such as W_ _ _ _ P _ G,  _ _ _ _ _ W P G, etc., as equal to the number of unique locations in the word for W,P,G while preserving their order.  This alone comes out to ₈C₅ = 56, which already overshoots your answer of 30, and is an underestimate given that the number of distinguishable permutations of letters filling the blanks hasn't yet been taken into account.

Answer 2

WINNIPEG

W, P, and G must remain in the same place

W_ _ _ _P_G

there are 5 choices for place 2 (right after W)

there are 4 choices for place 3

there are 3 choices for place 4

there are 2 choices for place 5

there is 1 choice for place 6 (between P and G)

it seems that there are 5!=5*4*3*2*1 choices for the empty spaces but there are two l's and two N's

there are permutations of n things where P₁ are the same, P₂ are the same, and so on

the formula is n!/P₁!*P₂!*P₃!*...*P_r!

5!/(2!*2!)=(5*4*3*2*1)/[(2*1)*(2*1)]=120/4=30 ways the letters can be arranged if W, P, and G must stay in the same order