CAT Practice: Permutation & Combination - 2 - Free MCQ Test with solutions

Let's take the number of paths between Q and R to be b and the number of paths between R and S to be a

We are given the paths from P to S through R (which would be 4a), the paths from P to S through Q (which would be 12) and the paths from P to Q to R to S, which would be 3ab) is equal to 62
Giving the relation 4a+12+3ab = 62
Or 4a+3ab = 50

The paths from Q to R directly (which would be b), through P( which would be 12) and through S (which would be 4a) are 27
Giving the relation b+12+4a = 27
Or 4a+b = 15

Subtracting this equation from the first one we got, we get 3ab-b=35, or b(3a-1)=35
b can be 1, 3, 5 or 7
Substituting these values in the second equation, we see that it can not be 1 or 5, leaving only 3 or 7 as the possible values. 

Substituting b as 3 in the first equation would give 13a=50, which is not true. 
Substituting bas 7 in the first equation would give 25a= 50, which would give a=2

We are asked the number of paths from Q to R, which is b=7

Therefore, 7 is the correct answer. 

Basically, any 4 digits chosen from 1-9, can be arranged in exactly 1 way, such that all four digits are in ascending order. Hence total number of such numbers possible is ⁹C₄ = 126.
For instance if the digits chosen are 7, 3, 4 and 8, we can arrange them in ascending order in exactly one way, as 3478.
Thus 126 such numbers are obtainable.

ways and the direction can be selected in 2 ways. So, total number of ways is 12 x 11 = 132

Case 1: When 7 is at the left extreme In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0, 1, 2, 4, 5, 6, 8, 9)

So total ways 3(8)(7)= 168

Case 2: When 7 is not at the extremes Here there are 3 cases possible. And the remaining two places can be filled in 7(7) ways.(Remember 0 can't come on the extreme left)

Hence in total 3(7)(7) = 147 ways

Total ways 168 + 147 = 315 ways

I. 8 consecutive Tails → 1 way

II. 6 Tails, 1 Head → 7C1 = 7 ways

III. 4 Tails, 2 Head → 6C2 = 15 ways

IV. 2 Tails, 3Head → 5C3 = 10 ways

V. 4 consecutive Heads → 1 way

Thus total possible number of ways = (1 + 7 + 15 + 10 + 1) = 34 ways.

This question is an application of the product rule in probability and combinatorics.
In the product rule, if two events A and B can occur in x and y ways, and for an event E, both events A and B need to take place, the number of ways that E can occur is xy. This can be expanded to 3 or more events as well.
Event 1: Distribution of balloons Since each child gets at least 4 balloons, we will initially allocate these 4 balloons to each of them.
So we are left with 15 - 4 x 3 = 15 - 12 = 3 balloons and 3 children.
Now we need to distribute 3 identical balloons to 3 children.

Event 2: Distribution of pencils
Since each child gets at least one pencil, we will allocate 1 pencil to each child. We are now left with 6 - 3 = 3 pencils.
We now need to distribute 3 identical pencils to 3 children.

Event 3: Distribution of erasers
We need to distribute 3 identical erasers to 3 children.

Applying the product rule, we get the total number of ways = 10 x 10 x 10 = 1000.

Let us first calculate the number of 4-digit numbers with at least one 8 and no 9.

We calculate all 4-digit numbers having no 9 and subtract from it the number of 4-digit numbers with no 8 and no 9 to get the required count.

Number of all 4-digit numbers having no 9 = 8 x 9 x 9 x 9

Number of 4-digit numbers with no 8 and no 9 = 7 x 8 x 8 x 8

Hence, the required count = 8 x 9 x 9 x 9 - 7 x 8 x 8 x 8 = 2248

To calculate the number of 4-digit numbers with at least one 9 and no 8, we will do the same and get 2248.

Hence, total = 4496.

Total number of different numbers of 4 digits formed by using these digits only once = 4!/2! = 12.

The numbers formed whose unit digit is 3 = 3

The number formed whose unit digit is 5 = 6

The number formed whose unit digit is 6 = 3

(Same rule is applicable for 10th, 100th and 1000th units)

Sum of the total number of unit digits = 3×3+5×6 +6×3 = 57

Sum of the 10th digit number = 57×10

Sum of the 100th digit number = 57×100

Sum of the 1000th digit number = 57×1000

Sum of all the numbers formed = 57(1+10+100 + 1000) = 63327.

So, India can win the series in 1+4+10+20 = 35 ways.
Similarly, Australia can win the series in 35 ways.
Hence, the total number of ways in which the series can be won is 35+35=70 ways.

The digits which can be recognised as unique digits when they are Inverted in a calculator are 0, 1, 2, 5, 6, and 9 Since the number cannot begin with zero all the numbers having 0 at units place should be discarded for otherwise when reed upside down the number yen begin with 3 we now :1St the different posibililities

Thus, the number of required numbers = 7 + 6² + 6² x 7 + .. + 6² x 7⁴