How many numbers of 7 digits can be formed from the digits 1, 2, 3, 4,

Question

How many numbers of 7 digits can be formed from the digits 1, 2, 3, 4, 5, 6, 7 if there are not more than 2 digits between 1 and 2 ?

A. 2160
B. 2400
C. 2640
D. 3120
E. 3600

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TestPrepUnlimited · Answer

First, we can isolate digits 1 and 2, and confirm how many scenarios we can have after slotting 1 and 2. We may have 12, 21, 1?2, 2?1, 1??2, or 2??1 as part of the 7 digits.

Next, we need to place these strings somewhere within the 7 slots _ _ _ _ _ _ _ . The rest of the slots will be filled in with 3 through 7. So we count how many positions we can place these strings, 12 and 21 have 6 positions they can fill in, 1?2 and 2?1 have 5 possible positions, and 1??2, 2???1 have 4 possible positions. In total, we have 12 + 10 + 8 = 30 positions for 1 and 2.

Finally, we fill in the rest of the blanks, for each position after slotting in 1 and 2, we have 5 blanks and 5 digits to choose from. Thus 5*4*3*2*1 = 120 numbers after we slot in 1 and 2. Therefore in total 30*120 = 3600 numbers.

Ans: E

CrackverbalGMAT · Answer

Case 1: No number between 1 and 2

We can have 12_ _ _ _ _ or _ 1 2 _ _ _ _ or _ _ 1 2 _ _ _ or _ _ _ 1 2 _ _ or _ _ _ _ 1 2 _ or _ _ _ _ _ 1 2

There are 6 ways and 1 and 2 can interchange in 2! = 2 ways. The remaining 5 numbers are arranged in 5! = 120 ways.

Total ways for the above = 6 * 2 * 120 = 1440

Case 2: One number between 1 and 2

We can have 1 _ 2 _ _ _ _ or _ 1 _ 2 _ _ _ or _ _ 1 _ 2 _ _ or _ _ _ 1 _ 2 _ or _ _ _ _ 1 _ 2

There are 5 ways and 1 and 2 can interchange in 2! = 2 ways. The remaining 5 numbers are arranged in 5! = 120 ways.

Total ways for the above = 5 * 2 * 120 = 1200

Case 3: Two numbers between 1 and 2

We can have 1 _ _ 2 _ _ _ or _ 1 _ _ 2 _ _ or _ _ 1 _ _ 2 _ or _ _ _ 1 _ _ 2

There are 4 ways and 1 and 2 can interchange in 2! = 2 ways. The remaining 5 numbers are arranged in 5! = 120 ways.

Total ways for the above = 4 * 2 * 120 = 960

Therefore Total number of ways = 1440 + 1200 + 960 = 3600

Option E

Arun Kumar

rsrighosh · Answer

CrackVerbalGMAT

For the 3rd case:- why aren't we taking 5C2 for the numbers between 1&2 ---> 1 _ _ 2 _ _ _ .
Why are we taking 5C1 * 4C1.

What is the difference between 5C1 * 4C1 and 5C2 while choosing numbers

CrackverbalGMAT · Answer

Hi rsrighosh. we know that there are 8 ways of arranging 1 and 2.

I can have 1 3 4 2 5 6 7. Now with 1 3 4 2 as the first 4 numbers, I can have 6 different numbers as the last 3 digits can arrange themselves in 3! ways.

Similarly I can have 3 5, 3 6, 3 7, 4 5, 4 6, 47, 5 6, 57 and 6 7. Total ways of having 2 numbers between 1 and 2 is 10, and with each of these i can have 6 numbers. So total numbers = 10 * 6 = 60. I can also have 10 ways where the numbers are written in reverse order like 4 3, 5 3 63.. and so on.

So total possibilities = 60 + 60 = 120

Now, when I take 5C2 = 10 ways, I am effectively saying that order does not matter, and that 1 3 4 2 is the same as 1 4 3 2. To nullify that effect, we must multiply 5C2 by 2! to ensure that both numbers are possible.

When taking it individually as 5C1 * 4C1, we are then ensuring both possibilities. i.e the first number can be any of 3 4 5 6 7 and the second number will be any of the 4 remaining. Here then 3 4 and 4 3 are taken into account.
In this case we do not need to multiply by 2! as we are choosing 1 at a time.

Hope this helps

Arun Kumar

PrecisionPrep · Answer

Alternatively, we can solve this one by counting the total possible numbers (ignoring the restriction) and then subtracting the number of cases where the restriction is violated.

okHedwig · Answer

My solution is like : (5!) (3+4+5+6+5+4+3) = 3600
so if we leave 1& 2, 5! ways we can place other digits.
Then placing different possible places for 1 & 2 like :
if 1 is 1st place, 3 ways 2 can be placed
if 1 is sec place, 4 ways 2 can be placed
if 1 is 3rd place, 5 ways 2 can be placed and so on.. till 1 in 7th place.

Hope its understandable. (1 Query though, it doesn't mention in the Question that nos. are non repeating!)

How many numbers of 7 digits can be formed from the digits 1, 2, 3, 4,

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How many numbers of 7 digits can be formed from the digits 1, 2, 3, 4,

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