pemutation combination

Ans 2) (a) 216

The sum of the digits of any number that is divisible by '3' is divisible by 3.

For instance, take the number 54372.
Sum of its digits is 5 + 4 + 3 + 7 + 2 = 21.
As 21 is divisible by '3', 54372 is also divisible by 3.

There are six digits viz., 0, 1, 2, 3, 4 and 5. To form 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits.

The sum of all the six digits 0, 1, 2, 3, 4 and 5 is 15. We know that any number is divisible by 3 if and only if the sum of its digits are divisible by '3'.

Combining the two criteria that we use only 5 of the 6 digits and pick them in such a way that the sum is divisible by 3, we should not use either '0' or '3' while forming the five digit numbers.

Case 1
If we do not use '0', then the remaining 5 digits can be arranged in 5! ways = 120 numbers.

Case 2
If we do not use '3', then the arrangements should take into account that '0' cannot be the first digit as a 5-digit number will not start with '0'.

. The first digit from the left can be any of the 4 digits 1, 2, 4 or 5.

Then the remaining 4 digits including '0' can be arranged in the other 4 places in 4! ways.

So, there will be 4*4! numbers = 4*24 = 96 numbers.

Combining Case 1 and Case 2, there are a total of 120 + 96 = 216 5 digit numbers divisible by '3' that can be formed using the digits 0 to 5.

[1]

Thank you very much,Khyati!!

Folks,no need to do the third one.I have found the answer.It is a)4.
Plz help me with the first question.

3)Divisors of 240 = 1, 2, 3, 4, 6, 8, 12, 16, 25, 5, 10, 20, 24, 48, 15, 30, 60, 120, 240, 80

From these divisors just 4 divisors 2, 6, 10 and 30 are of 4n+2 form, thus the answer is 4

1.it can as broken into as a+b=10
so pairs possible are:(1,9)(2,8)(3,7)(4,6)

similarly for c+d=10
now u may select 1 pair for a+b=10 out of 4
and 1 pair fr c+d=10 frm remaining 3

ie 4C1 x 3C1
and araange them in 4! ways =4 x 3 x 24 =288

Hello there

I like yur idea but this can not the answer as there are 2 reasons:

1.why only a+b=10 and c+d=10? I can also restrict the equation to a+b=3 and c+d=17,etc and so on

2.By using the restriction a+b=10 and c+d=10,you have completely omitted the usage of 5.Well, I can also write the equation as 5+9+4+2=20
where 5 can be used.

Taking these factors into account,can you plz help me with the question?

A not very efficient method that does the job is to look at all positive solutions to the equation

x_1+x_2+x_3+x_4 = 20 and weed out solutions where two or more variables are equal.

The number of +ve solutions is \binom{19}{3}

From this exclude cases where:
all the variables are equal - 1

3 are equal - 20

2 are equal - 26 (if my count is right)

So that makes the number of solutions = 969-47 = 922

My count was wrong for the case when two variables are equal (i got the count of distinct cases wrong and forgot to provide for permutations). So that case will have 32X12 = 384 instances

So the number of distinct solutions = 969-405 = 564

I dont get it.How come the number of positive solutions is C₃¹⁹?And plz explain the later half of yur message.

The number of positive integral solutions to x_1+x_2+...+x_r = n is given by \binom {n-1}{r-1}

no formula is required for 2nd one....just apply common sense......
the sum=0+1+2+3+4+5=15

to be divisible by 3, the sum has to be divisible by 3 right??

15 is already div. by 3!!

so the cases are:-

1) using-1,2,3,4,5
2)using-0,1,2,4,5 (eliminating 3 as 15-3=12...div. by 3)

thus the answer is 216!!