Back At It With PnC

1)

ac ad 

bc bd

ca cb 

da db 

8 arrangements 

1)  Try this:

4! - [4C0 + 4C1 + 4C2 + 4C3 + 4C4] = 24 - [ 1 + 4 + 6 + 4 + 1] = 24 - 16 = 8 arrangements.

4)  This one would be ALL multiples of 5 from 5 to 995 =995 / 5 = 199 - the repeated digits such as 55.

So, I count only one repeated digit from 5 to 99. And from 100 to 199, you have 100, 110, 115, 155, or 4 numbers. So, that pattern should hold all the way up to 995. So, you would have: 1 + (4 x 9) = 37 numbers, which have repeated digits in them.

So, 199 - 37 = 162 numbers that are divisible by 5 that have no repeated digits. You may check my numbers because I may have missed something.

2)  OK, I'm going to "guess" on this one. Since you have 10 digits to choose from, 0 to 9, then we have: Permutations of 10 digits taken 9 at a time: 10P9 =3,628,800 / 9 =403,200 that are divisible by 9 and that should not have any repeats in them. So, the value of K in this case would be:

3,628,800 / K8! and K = 9 x 10 = 90. Somebody should check this!!!.

3)  

This one would consist of all all numbers between 10,000 and 99,999.

So, we have: 99,999 - 10,000 + 1 =90,000 5-digit numbers. So, 90,000/2 - 1 =44,999, 5-digit numbers whose sum would be even and 45,001 whose sum would be odd because the first, 10,000 and the last, 99,999 are odd. Such as: 10,000 odd, 10,001 even, 10,002 odd, 10,003 even, 10,004 odd, 10,005 even......and so on to 99,998 even, 99,999 odd.

Somebody should check this one as well !!.

i know you all might be bored by now with this but i cant help it...ive got some more PnC questions...and there are more like these so if anyone could help thatd be great!

1. Find the no. of arrangements of the letters 'abcd' in which neither a,b nor c,d come together is? 

I will assume they will be in a row.

There are 4! arrangments altogether.  4*3*2*1 = 24

In how many are they together..

I just counted

So I think that is 16

24-16=8

So that leaves 8 permutations where A is not next to B and C is not next to D

So this one agrees wit the other answerers  

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2. No. of 9 digit nos. divisible by 9 using the digits from 0 to 9 if each digit is used atmost once is K.8! , then find the K has the value equal to_______.

The digits will have to add to a multiple of 9

1,2,3,4,5,6,7,8,9 add to 45            9! =  9*8! possibilities

0,1,2,3,4,5,6,7,8 add to 36            8 possibles for the highest vaue digit (can't be zero) * 8! 

these are the only 2 possible sets of digits. 

So that is  9*8! + 8*8! = 17*8!

K=17

3. Fine the no. of 5 digit nos. such that the sum of their digits is even is.

You will need an even number of odd digits for this to work.

So All even, or 3 even or 1 even  (the other digits odd)

All even. the biggest canonot be 0.  4*5^4 = 2500

3 even and 2 odd  

e e e o o     there are 5C3 ways this can be arranged 

How many ways are there to get 3 even digits from 5 where order counts and repetition is allowed. mmm

I think that is  5^3

How many ways are there to get 2 odd digits from 5 where order counts and repetition is allowed. 

I think that is  5^2

So that seems to be     5^3*5^2*5C3 = 5^5 * 5C3 = 3125*10 = 31250 ways

But some of those have the biggest digit 0 and that is not allowed.   

How many have the first digit =0???

5^2*5^2* 4C2 = 5^4 * 4C2 = 625 * 6 = 3750

so 

31250-3750 = 27500

I think there are 27500 five digit numbers with 3 even digits and 2 odd digits

How many numbers are there with 1 even digit and 4 odd ones.

0 followed by 4 odds ... there are    5^4 = 625 (these are NOT included)

There are now 5 even ones to chose from and this can go 1st, 2nd, 3rd, 4th or 5th that is 5 places.

So that is 25 possibilities for what even one goes where.

Now there are 5^4 ways to chose the odds = 625 ways

Altogether there are  625*25 = 15625 numbers with 4 odd and 1 even digit but I have to subtract the ones that start in 0

15625 - 625 = 15000

I think there are 15000 five digit numbers with 1 even digits and 4 odd digits

So altogether that makes  2500+27500+15000 = 45000 ways.

Just like the others have said, I am not certain this is correct. 

4. Find the no. of natural nos. less than 1000 and divisble by 5 can be formed with the ten digits, each digit not occuring more than once in each number is______.

If the units is 0 then there are 9 choices for the 10s and 8 choices for the hundreds =     72 that end in zero

If the units is 5 and the tens is 0 then there are 8  choices for the hundreds =                    8 ending in 05

If the units is 5 and the tens is NOT 0 then there are 7  choices for the hundreds = 8*7 = 56  ending in not0 5

So that seems to be   72+8+56 = 136 numbers fit the criterion

Rosala, when you get the answers it would be good if you share them with us.

Make a new post that directs us to this one so that we will see it please