Permutations & Combinations – III

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1 2 3 4 5 6
The black flags will be placed on either even places (2, 4, 6) or odd places (1, 3, 5)

Alexa can first choose a white or a black square in 32 ways (as there are 32 black and 32 white squares).
Now, she can choose the square of other column which does not lie in the same row or column (as that of the square chosen above) in 24 ways.
Total number of ways = 32*24 = 768

The total number of solutions depends on the number of ways in which 30 can be resolved into three factors, and further on the number of ways in which these respective factors can be arranged.

Now, 30 = 2*3*5
2, 3 and 5 can be arranged in 3 * 2* 1 = 6 ways

Similarly, 30 can be resolved in the following other ways:

(1*2*15), (1*3*10) and (1*5*6). For each of these ways, the factors can be arranged in 6 ways (as above).

Further, 30= 1*1*30
These 3 factors can be arranged in 3! /2! = 3 ways (as two factors are the same)

Total = 6+18+3 = 27

Total number of different 4 digits formed by using three digits once = 4!/2! = 12
The numbers formed whose unit’s digit is 3 = 3
The numbers formed whose unit’s digit is 5 = 6
The numbers formed whose unit’s digit is 6 = 3
(same rule applicable for 10^th, 100^th and 1000^th units)
Sum of the total number of units digit = (3*3)+(5*6)+(6*3) = 57
Sum of 10^th digits = 57*10
Sum of 100^th digits = 57*100
Sum of 1000^th digits = 57*1000
Therefore, sum = 57(1+10+100+1000) = 63327

12 students can arrange themselves in 12! ways.
The 3 sticks, 2 of the same colour can be arranged in 3! /2! = 3 ways.
-> Total number of required ways = 12! *3

Since the first pencil can be given to any of the 5 students, so the number of ways in which the first pencil can be given away = 5

In a similar manner, the second, third, fourth, fifth and sixth pencils can be given away in 5 ways each.

-> The required number of ways = 5⁶ = 15625

For each of the 10 questions, a student can either omit it or do part A or part B. Hence 03 ways to tackle each question exist.
For 10 questions we have 3¹⁰ways, Finally, subtract 1 to avoid the possibility of omitting all.
Required answer = 3¹⁰ – 1 = 59048

Total no. of numbers = 4*4*3*2*1 = 96 (because 1st place can’t be occupied by 0)
Divisibility by 75 means divisible by both 3 and 25 both.
All numbers formed with the given digits are divisible by 3 (because sum is 18).
For divisibility by 25 the rule is that the last two digits of the number should form a multiple of 25 i.e., 25, 75 or 50.
If last 02 digits are 25 then 2*2*1|2 5 -> there are 4 such numbers
If last 02 digits are 75 then 2*2*1|7 5 -> there are 4 such numbers
If last 02 digits are 50 then 3*2*1|5 0 -> there are 6 such numbers
Total numbers divisible by 25 = 4+4+6 = 14
Answer = 96-14 = 82

1-digit number will be 8 only i.e., 1
2-digit numbers will be such that the last two digits should form a multiple of 4 i.e., these numbers will end in 56, 68, 76, 88, 96 -> there will be 5 such numbers
3-digit numbers will be such that the last two digits can be occupied in 5 ways only, as these numbers have to end in 56, 68, 76, 88 or 96; and the first digit (hundredth place) can be occupied in 5 ways -> total number of such numbers = 5*5 =25
Now, 4-digit numbers = 5*5*5 = 125
5-digit numbers = 5*5*5*5 = 625
Total = 625+125+25+5+1 = 781

Total numbers of 4-digit numbers = 6⁴
Out of these, in every set of 6 numbers, we have 2 numbers divisible by 3, as shown below:

Now, in every 6 numbers, 2 numbers are divisible by 3
-> In every 6⁴ numbers, those divisible by 3 = (2/6) *6⁴ = 2*6³ = 432