# How many n-digit numbers can be formed using the first 5 natural numbers which must contain the digits 2 and 4 essentially [closed]

As the title says, how many n-digit numbers can be formed using the first 5 natural numbers which must contain the digits 2 and 4 essentially?

I think this can be done using some sort of principle of inclusions, but I am not able to do it in that way...

Can someone guide me in the right direction?

## closed as off-topic by JMoravitz, Arnaud D., Shailesh, Marcus M, AquaNov 8 '17 at 17:32

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• Do you consider $0$ a natural number? Yes the problem will utilize the principle of inclusion-exclusion. Can you count the number of such $n$-digit numbers with no twos? Can you count the number of such $n$-digit numbers with no fours? Can you count the number of such $n$-digit numbers with neither twos nor fours? Can you count the number of such $n$-digit numbers where you don't care about whether there are any of either? – JMoravitz Nov 8 '17 at 5:39
• @JMoravitz 0 is not considered as a natural number in this case, as usual, and thanks for the insight... That helped me to take the right cases... – AbhigyanC Nov 8 '17 at 16:09
• "is not considered as a natural number, as usual" I find it is much more common to consider zero a natural than otherwise, but both are used. it feels like maybe 70-30 in favor of zero being considered a natural – JMoravitz Nov 8 '17 at 16:10
• @JMoravitz Oh... I was unaware that 0 is even treated as a natural number in any place, having grown up hearing that there's natural numbers ($\{1,2,3,4...\}$) and whole numbers ($\{0,1,2,3,4...\}$)... I'm from India, here we have always learnt that 0 is a natural number; hence I said "as usual" :) – AbhigyanC Nov 8 '17 at 17:04

Let us consider multiple cases to solve this:

Total number of numbers present of n-digit made up of 1st 5 natural no.s is $S=5^n$

Case 0: This case considers with atleast one 2 present and no 4's present

$\implies S_0= \ ^nC_13^{n-1} +\ ^nC_23^{n-2} +\ ^nC_33^{n-3}+ \ ...+ \ ^nC_n$

$\implies S_0=4^n-1$

Case 1: This case considers with atleast one 4 present and no 2's present

$\implies S_1= \ ^nC_13^{n-1} +\ ^nC_23^{n-2} +\ ^nC_33^{n-3}+ \ ...+ \ ^nC_n$

$\implies S_1=4^n-1$

Case 2: No 2's or 4's are present

$\implies S_2=3^n$

So, total no. of such values are

$s=S-S_0-S_1-S_2$

$\implies s=5^n-2(4^n-1)-3^n$