Find the number of 8-digit numbers the sum of whose digits is 4.
I am confused as I got the result as 120 and some of my friends told me they got 149.
Is their any formula to find it.
friendship is tan 90°.
You are correct, 120 is the answer.
The answer is done using generating functions but can also be done playing spot the pattern.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Thinking is cheating.
The possible combinations of nonzero digits in such an 8-digit number are as follows:
We take each case in turn.
(i) The only possible 8-digit number is 40000000.
(ii) The leading digit must be 1 or 3, and the other digit can be placed in any of the other 7 places. Thus there are 7 + 7 = 14 such 8-digit numbers.
(iii) One 2 is the leading digit and the other 2 can be placed in any of the other 7 places, so number of such 8-digit numbers is 7.
(iv) If the leading digit is 2, the two 1s can be placed in the other places in [sup]7[/sup]C[sub]2[/sub] = 21 ways. If the leading digit is 1, the other two digits can be placed in the other places in [sup]7[/sup]P[sub]2[/sub] = 42 ways. ∴ Number of such 8-digit numbers = 21 + 42 = 63.
(v) One of the 1s is the leading digit and the other 3 can be placed in the other places in [sup]7[/sup]C[sub]3[/sub] = 35 ways.
Hence the total number of such 8-digit numbers is 1 + 14 + 7 + 63 + 35 = 120.
Last edited by Nehushtan (2013-12-07 22:06:35)
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