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LMP
Joined: 29 Jun 2018
Last visit: 15 Mar 2022
Posts: 49
Given Kudos: 62
Location: India
Schools: Johnson '20 HEC Montreal '21
GPA: 4
01 Aug 2019, 07:16
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
38% (01:52) correct
62%
(02:08)
wrong
based on 63
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History
Date
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How many 8 digit cell phone numbers are there in which the digits do not increase from left to right ?
A) 17!/(8!9!)
B) 15!/(8!9!)
C) 12!/(9!8!)
D) 17!/(6!8!)
E) 12!/(8!7!)
Source: Private tutor
A) 17!/(8!9!)
B) 15!/(8!9!)
C) 12!/(9!8!)
D) 17!/(6!8!)
E) 12!/(8!7!)
Source: Private tutor
A bit tedious approach tho
⦁ There is only 1 way when phone number starts with '0'; That is '00000000'
⦁ Number of ways when phone number starts with 1 = n=8 (11111111, 11111110...........10000000)
⦁ Number of ways when phone number starts with 2 = Σn= n*(n+1)/2= 8*9/2=36
⦁ Number of ways when phone number starts with 3 = Σn*(n+1)/2= 120
⦁ Number of ways when phone number starts with 4 = Σ Σn*(n+1)/2= 330
We can see the pattern on pascal triangle that is similar to our above calculation
Total phone numbers possible= 1+8+36+120+330+792+1716+3432+6435+11440= 24310
⦁ There is only 1 way when phone number starts with '0'; That is '00000000'
⦁ Number of ways when phone number starts with 1 = n=8 (11111111, 11111110...........10000000)
⦁ Number of ways when phone number starts with 2 = Σn= n*(n+1)/2= 8*9/2=36
⦁ Number of ways when phone number starts with 3 = Σn*(n+1)/2= 120
⦁ Number of ways when phone number starts with 4 = Σ Σn*(n+1)/2= 330
We can see the pattern on pascal triangle that is similar to our above calculation
Total phone numbers possible= 1+8+36+120+330+792+1716+3432+6435+11440= 24310
LMP
How many 8 digit cell phone numbers are there in which the digits do not increase from left to right ?
A) 17!/(8!9!)
B) 15!/(8!9!)
C) 12!/(9!8!)
D) 17!/(6!8!)
E) 12!/(8!7!)
Source: Private tutor
A) 17!/(8!9!)
B) 15!/(8!9!)
C) 12!/(9!8!)
D) 17!/(6!8!)
E) 12!/(8!7!)
Source: Private tutor
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01 Aug 2019, 13:53
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It's not clear what the question means - you could interpret it in one of two ways:
- the question could be asking "how many phone numbers can we make where the digits do not strictly increase from left to right?" As the question is worded, that's how I'd interpret its meaning. Then we'd be counting almost every phone number -- 89731224 for example, does not strictly increase from left to right, since it sometimes goes up and sometimes goes down. The only numbers we would not count are numbers like 12345689 and 23456789. Assuming we can use any digit anywhere (so the number can start with zero - the question really should tell you if there are any restrictions like that, because some people will assume there are), we'd then have 10 choices for each digit, and 10^8 possible phone numbers in total, strictly increasing or not. We'd then need to exclude all of the strictly increasing phone numbers. But if we just pick two numbers from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and erase them, we'll be left with a string of eight digits in increasing order, so there are 10C2 = (10)(9)/2! = 45 strictly increasing numbers, and 10^8 - 45 that are not strictly increasing.
- but the question could be asking "how many phone numbers can we make where each digit after the first is either less than or equal to the digit before it?", or in other words, phone numbers with digits that are "constantly non-increasing". Then we'd be counting numbers like 99987100 and 65433331, because from one digit to the next, these numbers never go up - they either stay the same or go down. From the answer choices, this is obviously the meaning your tutor intends. You can use a 'partitions' method to answer questions like this (though if you're preparing for the GMAT, you might ask your tutor to show you even one official problem where you need to use partitions to get an answer). You can also look at the problem in this way: if ABCDEFGH is a constantly non-increasing string of eight digits between 9 and 0, then A, B-1, C-2, D-3, E-4, F-5, G-6, H-7 would need to be a strictly decreasing string of eight integers between 9 and -7 (and vice versa - from each decreasing sequence like that, we can create a unique phone number). So we just need to count how many strictly decreasing sequences we can make of eight integers between 9 and -7 inclusive, so we just need to choose 8 integers from the 17 in that range , and the answer is 17C8 = 17! / (8!)(9!).
You won't need to do this kind of thing on the GMAT though.
- the question could be asking "how many phone numbers can we make where the digits do not strictly increase from left to right?" As the question is worded, that's how I'd interpret its meaning. Then we'd be counting almost every phone number -- 89731224 for example, does not strictly increase from left to right, since it sometimes goes up and sometimes goes down. The only numbers we would not count are numbers like 12345689 and 23456789. Assuming we can use any digit anywhere (so the number can start with zero - the question really should tell you if there are any restrictions like that, because some people will assume there are), we'd then have 10 choices for each digit, and 10^8 possible phone numbers in total, strictly increasing or not. We'd then need to exclude all of the strictly increasing phone numbers. But if we just pick two numbers from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and erase them, we'll be left with a string of eight digits in increasing order, so there are 10C2 = (10)(9)/2! = 45 strictly increasing numbers, and 10^8 - 45 that are not strictly increasing.
- but the question could be asking "how many phone numbers can we make where each digit after the first is either less than or equal to the digit before it?", or in other words, phone numbers with digits that are "constantly non-increasing". Then we'd be counting numbers like 99987100 and 65433331, because from one digit to the next, these numbers never go up - they either stay the same or go down. From the answer choices, this is obviously the meaning your tutor intends. You can use a 'partitions' method to answer questions like this (though if you're preparing for the GMAT, you might ask your tutor to show you even one official problem where you need to use partitions to get an answer). You can also look at the problem in this way: if ABCDEFGH is a constantly non-increasing string of eight digits between 9 and 0, then A, B-1, C-2, D-3, E-4, F-5, G-6, H-7 would need to be a strictly decreasing string of eight integers between 9 and -7 (and vice versa - from each decreasing sequence like that, we can create a unique phone number). So we just need to count how many strictly decreasing sequences we can make of eight integers between 9 and -7 inclusive, so we just need to choose 8 integers from the 17 in that range , and the answer is 17C8 = 17! / (8!)(9!).
You won't need to do this kind of thing on the GMAT though.
