In how many ways can the letter of the word

below I've wrote letters without repeated words and then how many times particular word repeated

from the question here we gonna select 4 words hence we have the following chance to select those 4 words

Hence

**PROPORTION**be arranged by taking 4 letters at a time?**Solution:**

**P R O P O R T I O N**below I've wrote letters without repeated words and then how many times particular word repeated

**P R O T I N****P R O****O**

**Here (P, P) 2 P, (R, R) 2 R, and (O,O,O) 3 O**from the question here we gonna select 4 words hence we have the following chance to select those 4 words

**Chance 1:**3 O and remaining 1 are different (That means (O,O,O)--> 3 word and (P,R,T,I,N) ---> 1 word from the 5)**Chance 2:**two of same words and other two of the same words (here (O,O,O), (T,T) and (P,P) out of 3 pairs we gonna select 2 pair of words )**Chance 3:**Two of same words and other two are different words (here (O,O,O), (T,T) and (P,P) out of 3 pairs we gonna select 1 pair of word and other 2 word from (R, O, I, N, T) words consider P,P has been selected pair )**Chance 4:**All 4 are different ( here P, R, O, T, I , N out of 6 we gonna select 4 words)Hence

**Chance 1: 3C3 * 5C1 * (4! /3!) = 20 [3! for repeated words O,O,O]**

Chance 2: 3C2 * (4! / (2!*2!)) = 18 [2! * 2! for repeated words of T,T and P,P]

Chance 3: 3C1*5C2 * (4! / 2!) = 360 [2! for repeated word T,T or P,P]

Chance 4: 6C4 * 4! = 360 [as usual]

finally add all the chances = 20 + 18 + 360 + 360 = 758 ways is the answerChance 2: 3C2 * (4! / (2!*2!)) = 18 [2! * 2! for repeated words of T,T and P,P]

Chance 3: 3C1*5C2 * (4! / 2!) = 360 [2! for repeated word T,T or P,P]

Chance 4: 6C4 * 4! = 360 [as usual]

finally add all the chances = 20 + 18 + 360 + 360 = 758 ways is the answer