On how many ways can the letters of the word "COMPUTER" be arranged?
1. M must always occur at the third place
2. Vowels occupy the even positions.
Soln: For 1.
7*6*1*5*4*3*2*1=5,040
For 2.) I think It should be 4 * 720
there are 4 even positions to be filled by three even numbers.
in 5*3*4*2*3*1*2*1 It is assumed that Last even place is NOT filled by a vowel. There can be total 4 ways to do that.
Hence 4 * 720
My question is regarding the second part of the question
where we are calculating the ways in which vowels can be arranged
shouldnt it be
n!/(n-r)! RESULTING IN 24*720 as the order does matter
rather than n!/(n-r)!*r! which results in 4*720 order doesnt matter
the order should matter as e.g abc is not the same as cba when finding ways to arrange words
can someone please help
thanks
combination permutation
This topic has expert replies
Adeel,
First off this seems like a data sufficiency question which means that you dont have to determine a final answer but merely sufficiency. That being sad i think stmnt 1 is insufficient by itself given the existence of statement two or in other words you cannot be sure of any other constraints (except M is to remain put in the third position).
Likewise stmnt two is insufficient as well. Hence the answer choice drills down to either C or E.
I came up with 144 combinations and C as an answer choice by thinking about it like this:
C-O-M-P-U-T-E-R (8 slots minus 1 since M cant be moved so 7 slots)
now for the first slot i have 4 possibilities (CPTR) since vowels must be in the even slots. Furthermore the next slot i dealt with was slot no. 5 (U) here i have 3 possibilities (since 1 has been used up in the first). continuing this methodology the odd slots (excluding M) can be arranged in 4*3*2*1 or 24 ways. we exlude M since it cant be moved.
The even slots can be arranged in 3*2*1 (or 6 ways). The total combinations thus are 24*6 or 144.
Hope this helps. Thanks.
First off this seems like a data sufficiency question which means that you dont have to determine a final answer but merely sufficiency. That being sad i think stmnt 1 is insufficient by itself given the existence of statement two or in other words you cannot be sure of any other constraints (except M is to remain put in the third position).
Likewise stmnt two is insufficient as well. Hence the answer choice drills down to either C or E.
I came up with 144 combinations and C as an answer choice by thinking about it like this:
C-O-M-P-U-T-E-R (8 slots minus 1 since M cant be moved so 7 slots)
now for the first slot i have 4 possibilities (CPTR) since vowels must be in the even slots. Furthermore the next slot i dealt with was slot no. 5 (U) here i have 3 possibilities (since 1 has been used up in the first). continuing this methodology the odd slots (excluding M) can be arranged in 4*3*2*1 or 24 ways. we exlude M since it cant be moved.
The even slots can be arranged in 3*2*1 (or 6 ways). The total combinations thus are 24*6 or 144.
Hope this helps. Thanks.
- PussInBoots
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I though 1 & 2 were both required at the same time. Here is what I got:
4C3 {4 even position, 3 vowels} * 1 {M is 3rd} * 4! = 96
I bet this question is not from official guides/prep, because it's stated poorly.
4C3 {4 even position, 3 vowels} * 1 {M is 3rd} * 4! = 96
I bet this question is not from official guides/prep, because it's stated poorly.