+0  
 
+1
61
3
avatar+4690 

Help 18

Using the letters in the word INNOVATIVE, find the number of permutations that can be formed using 4 letters at a time. Show your work or explain how you got your answer.

 Dec 12, 2018
 #1
avatar
+1

Using the letters in the word INNOVATIVE, find the number of permutations that can be formed using 4 letters at a time. Show your work or explain how you got your answer.

 

Melody to the rescue!!. My computer says there are 1,398 permutations! But, for the life of me, I have NO IDEA how it gets that odd number! If you could figure it out, that would be just great. Thanks.

 

This much I learned: 

There are 246 "words" that begin with "I"

There are 246 "words" that begin with "N"

There are 246 "words" that begin with "V"

There are 165 "words" that begin with "O, A, T, E" for each.

So: [3 x 246] + [4 x 165] = 1,398 "words"

 Dec 12, 2018
 #2
avatar+812 
0

INNOVATIVE has 10 letters, so for a 4 letter word, there is \({10}\choose{4} \) = 210 permutations without considering overcounting. We need to count every permutation ONCE and ONLY ONCE. But if your question does not consider overcounting, then the number of permutations with 4 letters (not necessarily words) is \(\boxed{210}\).

 

 

 

If the problem does consider overcounting, then the solution is different. We did overcount (The word \(N_1 OTE\) is the same as \(N_2OTE\)). There are 2 i's, 2 n's, and 2 v's, so to eliminate overcounted permutations, we divide by 210 or subtract some numbers (i am not very sure).\(\)

 Dec 13, 2018
edited by PartialMathematician  Dec 13, 2018
edited by PartialMathematician  Dec 13, 2018
 #3
avatar+3576 
+1

\(\text{The way to look at this semi formally is to consider the different cases}\\ \text{There can be 0-2 pairs of letters chosen}\\ \text{with 0 pairs there are }\dbinom{7}{4}4! =840 \text{ unique 4 letter words}\\ \text{with 1 pair there are }3\dbinom{4}{2}\dbinom{6}{2}2!=540 \text{ unique 4 letter words}\\ \text{This last bit deserves some discussion}\\ \text{The 3 comes from the fact that 3 of the letters can be chosen as a pair, I, N, V}\\ \text{The }\dbinom{4}{2} \text{ comes from the fact that we select two slots from the 4 to place this pair}\\ \text{The }\dbinom{6}{2} \text{ comes from the fact that we can select 2 letters from 6 to complete the word}\\ \text{The 6 comes from removing the numbers that are the pair, and the }\\ \text{2nd copies of the two other numbers}\\ \text{Finally the }2! \text{ comes from the fact the these last two letters are distinct and their order matters}\)

 

\(\text{Finally with 2 pairs there are }\dbinom{3}{2}\dbinom{4}{2} =18 \\ \text{This is choosing two pairs from the 3 available and choosing 2 slots from the 4}\\ \text{adding all these up we get}\\ 840+540+18 = 1398 \)

.
 Dec 13, 2018

25 Online Users

avatar
avatar
avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.