How many 8 digit cell phone numbers are there in which the digits do

Question

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

nick1816 · Answer

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

IanStewart · Answer

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.

DarkHorse2019 · Answer

Read the above 2 solutions but still unclear how to go about in the real test scenario

I would like to request moderators to throw some light on this  chetan2u ,  Gladiator59 ,  VeritasKarishma ,  Bunuel ,  generis ,  gmatbusters

chetansood · Answer

Need a proper OA and official explanation.

This seems way too tough.

Pascals triangle approach cant be expected from most of us and i cant think of any other solution.

Posted from my mobile device

Rakesh1987 · Answer

For solving the question we need to understand two things:
1)   Not in Ascending order doesn't mean in descending order.   In 55444331 digits are not in ascending order but not necessarily in descending as well for eg 55. another example, in number 88888888, the digits are not in ascending order
2)   This is not a permutation but a combination question.   After we select the digits there is only one possible combination in which the digits is not in ascending order. 55444331 cant be written in any other way as per constraints of the question

Now approaching the question.
1)   As order doesn't matter, we just need to select digits for our number.   So all digit spaces in the 8 digit number xxxxxxxx are of equal importance.
2)   We need to select from among 10 numbers from 0-9, therefore this question is basically asking us to divide the 8 digits xxxxxxxx into 10 spaces  , the right most is for 9 and left most is for 0, and the number of x which falls in the range of a digit (0-9) is the frequency of that digit in our final 8 digit number. An example would look like this ||xx|xxx||xx|x||| is 77666443.

First line can be drawn in 9 ways, second in 10 and 9th in 17 ways and line can arrange amongst themselves in 9! ways. therefore the answer is 17!/9!8!. A.
Or using formula (n+r-1)C(r-1)=17!/8!9! A, where n is objects to be distributed (x) and r is number of boxes/ here digits (lines is r-1).

How many 8 digit cell phone numbers are there in which the digits do

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How many 8 digit cell phone numbers are there in which the digits do

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