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A registration number is made using 3 letters of the English alphabets
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Updated on: 22 Jun 2019, 13:14
Question Stats:
25% (02:58) correct 75% (03:09) wrong based on 20 sessions
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A registration number is made using 3 letters of the English alphabet followed by three numerical digits. The letters can be in any order but the numbers should be in a nonincreasing order  the second digit is not greater than the first, and the third digit is not greater than the second digit. How many such registration numbers can be formed? A. \(26^3*240\) B. \(26^3*220\) C. \(26C3*240\) D. \(26C3*220\) E. \(26^3*320\)
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Originally posted by firas92 on 21 Jun 2019, 22:13.
Last edited by firas92 on 22 Jun 2019, 13:14, edited 1 time in total.



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Re: A registration number is made using 3 letters of the English alphabets
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21 Jun 2019, 23:11
Struggling with the statement: numbers should be in nonincreasing order. Can someone help me find such 3 digit numbers . The other part should obviously be 26^3



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Re: A registration number is made using 3 letters of the English alphabets
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22 Jun 2019, 07:55
We have 26 choices for each letter, so 26^3 choices for the first three letters. For the three nonincreasing digits, there are three possibilities:  the three digits are identical: then we have 10 possibilities (000, 111, 222, etc)  two of the digits are identical and one different. We then have 10 ways to choose the identical pair, and 9 remaining ways to choose the other digit, for 90 possible choices. Once we choose the three digits, there will only be one order we can put them in that will be nonincreasing, so we have 90 possibilities in this case.  all three digits are different. Then we have (10)(9)(8)/3! = 120 ways to choose the three digits, and once we've chosen the digits, only one order we can put them in will be nonincreasing. So we have 120 possibilities in this case. So there are 10+90+120 = 220 ways to make the number, and 26^3 * 220 registration numbers in total. There's a much faster way to choose how to count the three digits, but the method uses partitions in a complicated way, and so is well beyond GMAT level math.
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Re: A registration number is made using 3 letters of the English alphabets
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22 Jun 2019, 08:18
IanStewart could you please explain the highlighted part of your solution ; my query is while choosing a three digit no where 2 of the digits are identical so in that case shouldnt the combination be 10*1*9* 3 ; for eg ; 122, 212,221 we can get three different nos. the question though has stated that " the numbers should be in a nonincreasing order"
eg ; 571 , 481 etc. how to identify such no? is the approach of 10*1*9 ; 90 correct ? IanStewart wrote: We have 26 choices for each letter, so 26^3 choices for the first three letters.
For the three nonincreasing digits, there are three possibilities:
 the three digits are identical: then we have 10 possibilities (000, 111, 222, etc)  ttwo of the digits are identical and one different. We then have 10 ways to choose the identical pair, and 9 remaining ways to choose the other digit, for 90 possible choices. Once we choose the three digits, there will only be one order we can put them in that will be nonincreasing, so we have 90 possibilities in this case.  all three digits are different. Then we have (10)(9)(8)/3! = 120 ways to choose the three digits, and once we've chosen the digits, only one order we can put them in will be nonincreasing. So we have 120 possibilities in this case.
So there are 10+90+120 = 220 ways to make the number, and 26^3 * 220 registration numbers in total.
There's a much faster way to choose how to count the three digits, but the method uses partitions in a complicated way, and so is well beyond GMAT level math.



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Re: A registration number is made using 3 letters of the English alphabets
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22 Jun 2019, 08:42
Archit3110 wrote: my query is while choosing a three digit no where 2 of the digits are identical so in that case shouldnt the combination be 10*1*9* 3 ; for eg ; 122, 212,221 we can get three different nos. the question though has stated that "the numbers should be in a nonincreasing order" eg ; 571 , 481 etc. how to identify such no? is the approach of 10*1*9 ; 90 correct ?
I see  the wording of the question is ambiguous. In my solution, when they say "nonincreasing order", I've interpreted that to mean "the digits never increase". So in my solution, 841 is in nonincreasing order, but 481 is not  the digits increase from 4 to 8, then decrease from 8 to 1. I think you're interpreting "nonincreasing order" to mean "not in strictly increasing order". Under your interpretation, 481 is "nonincreasing", and so is 841, but 148 is not. If you interpret the question in that way, the answer will be much larger. The wording of the question simply isn't clear, so both of our interpretations make logical sense  there's no way to tell even what the question is asking us for, so no way to say what the "right" answer is. But under my interpretation, once you choose a pair of digits and a third different digit, which you can do in 90 ways, there is then only one neverincreasing order you can put the digits in. If you choose 2, 2, and 1, the only number you can make with neverincreasing digits is 221.
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Re: A registration number is made using 3 letters of the English alphabets
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22 Jun 2019, 10:03
yes you are correct IanStewart question wording is kinda ambiguous Hey firas92 could you please share the source of this question ? IanStewart wrote: Archit3110 wrote: my query is while choosing a three digit no where 2 of the digits are identical so in that case shouldnt the combination be 10*1*9* 3 ; for eg ; 122, 212,221 we can get three different nos. the question though has stated that "the numbers should be in a nonincreasing order" eg ; 571 , 481 etc. how to identify such no? is the approach of 10*1*9 ; 90 correct ?
I see  the wording of the question is ambiguous. In my solution, when they say "nonincreasing order", I've interpreted that to mean "the digits never increase". So in my solution, 841 is in nonincreasing order, but 481 is not  the digits increase from 4 to 8, then decrease from 8 to 1. I think you're interpreting "nonincreasing order" to mean "not in strictly increasing order". Under your interpretation, 481 is "nonincreasing", and so is 841, but 148 is not. If you interpret the question in that way, the answer will be much larger. The wording of the question simply isn't clear, so both of our interpretations make logical sense  there's no way to tell even what the question is asking us for, so no way to say what the "right" answer is. But under my interpretation, once you choose a pair of digits and a third different digit, which you can do in 90 ways, there is then only one neverincreasing order you can put the digits in. If you choose 2, 2, and 1, the only number you can make with neverincreasing digits is 221.



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Re: A registration number is made using 3 letters of the English alphabets
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22 Jun 2019, 10:15
Archit3110 wrote: yes you are correct IanStewart question wording is kinda ambiguous Hey firas92 could you please share the source of this question ? IanStewart wrote: Archit3110 wrote: my query is while choosing a three digit no where 2 of the digits are identical so in that case shouldnt the combination be 10*1*9* 3 ; for eg ; 122, 212,221 we can get three different nos. the question though has stated that "the numbers should be in a nonincreasing order" eg ; 571 , 481 etc. how to identify such no? is the approach of 10*1*9 ; 90 correct ?
I see  the wording of the question is ambiguous. In my solution, when they say "nonincreasing order", I've interpreted that to mean "the digits never increase". So in my solution, 841 is in nonincreasing order, but 481 is not  the digits increase from 4 to 8, then decrease from 8 to 1. I think you're interpreting "nonincreasing order" to mean "not in strictly increasing order". Under your interpretation, 481 is "nonincreasing", and so is 841, but 148 is not. If you interpret the question in that way, the answer will be much larger. The wording of the question simply isn't clear, so both of our interpretations make logical sense  there's no way to tell even what the question is asking us for, so no way to say what the "right" answer is. But under my interpretation, once you choose a pair of digits and a third different digit, which you can do in 90 ways, there is then only one neverincreasing order you can put the digits in. If you choose 2, 2, and 1, the only number you can make with neverincreasing digits is 221. Archit3110https://cracku.in/I guess this is not typically a source for GMAT questions.



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Re: A registration number is made using 3 letters of the English alphabets
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22 Jun 2019, 10:21
OFFICIAL EXPLANATION
The letters can be picked in 26*26*26 ways
A non increasing set of numbers means that a digit that follows another digit cannot be more in value than the preceding digit.
In order to select the numbers in a non increasing sequence,
If all three are distinct, the number of cases is 10C3=120
If exactly two of the three digits are the same, the number of ways of selecting the two digits is 10C2 each of which can be arranged in 2 ways. (Numbers like 221, 211 fall in this category). So this total is 10C2*2=90
The last case is if all three numbers were the same. There are 10 ways (000~999)
So the total number of ways of picking the digits is 120+90+10=220
Hence the number of possible registration numbers is \(26^3*220\)



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Re: A registration number is made using 3 letters of the English alphabets
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23 Jun 2019, 00:41
First 3 characters is a no brainer... 26^3 ways to do that
Last 3 char is tricky
Non increasing 3 digit codes
Possibility 1: all repeating digits =10
Possibility 2 : 2 digits repeating = 10*9 (10 for repeating and excluding the treated digits 1 digit remain for the other 3rd option)=90
Possibility 3: No repeating digits = (10*9*8)3! =120
Why divide by 3 factorial ..because there is only one way in which each paid of 3 digits can be (the non increasing way)
Total no of.codes = 26^3 *(10+90+120)
=26^3 * 220
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Re: A registration number is made using 3 letters of the English alphabets
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