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

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

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

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