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[Muhammad Talha]

Can anybody solve ques:Nov 05/P2/Q11(b).....!!!!
plz explain it also... :?:

 

[OakMoon!]

I think the marking scheme gives the perfect solution. To help you in solving this problem make 5 successive boxes. Since the first box can't have 0 we have to make a selection for the first number out of the 4 other possibilities which makes it 4C1=4.
Then permute the rest of the boxes. We have to arrange the 4 remaining numbers that include 0 in the remaining 4 boxes to do that we use 4!. Or 4P4. And then multiply them. 4!*4=96
For the second part we have to find the permutations for the different possible even numbers and then add them. A five digit even number will always have an even number or a 0 as its last digit. In this case we have 0,2 and 4. Find the different arrangements for each of the 3 numbers at the end of the 5 digits.
For 0 when the 0 is fixed at the last place the remaining four places can be filled by the 4 remaining numbers which makes it 4!=24.
When the last digit is 2. We know the first digit is either 1,3 or 4 and so we make a selection 3C1=3. The remaining 3 places include 0 and 2 other numbers which get the permutation of 3!. So for the even numbers ending with 2 we have the permutation 3*3!=18. And the same goes for the 5 digit even number ending with 4 so we double the figure as to add in both the values.
Final answer= 4!+2*3*3!=60

 

[leosco1995]

Hamidali gave a perfect explanation for this. I also drew a picture for this in case it helps a bit more. For part (ii), there are 3 cases for an even number: that the number ends in 0, 2 or 4 so I drew 3 boxes there.

 

[Muhammad Talha]

thanx to u both.....