Reliance Jio Permutation and Combination Quiz 1

correct answer is 32
Reason:-
The 5-letter word can be rearranged in 5!=120 Ways without any of the letters repeating.
The first 24 of these words will start with A.
Then the 25th word will start CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.
The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.
Therefore, the rank of CHASM will be 24+6+2= 32

Correct answer is 2880
Explanation:
There are a total of 9 places out of which 4 are even and the rest 5 places are odd.
4 females can be arranged at 4 even places in 4! ways. and 5 males can be placed in the remaining 5 places in 5! ways. Hence, the required number of permutations = 4! \times 5! = 24 \times 120 = 2880

The correct answer is 215
Explanation:
Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 \times 6 \times 6 = 216. Of these attempts, one of them is a successful attempt.
Maximum number of unsuccessful attempts = 216 - 1 = 215

Correct answer is 720
Explanation:
Let last digit is 2, When the second last digit is 4 the remaining 4 digits can be filled in 120 ways, similarly the second last digit is 6, the remaining 4 digits can be filled in 120 ways.
So for last digit = 2, total numbers=240
Similarly for 4 and 6
When last digit = 4, total no. of ways =240 and last digit = 6, total no. of ways =240
So a total of 720 even numbers are possible.

Correct answer is 2 \times (18!)
Explanation:
A person can be chosen out of 18 people in 18 ways to be seated between Mohan and Mohit. Now consider Mohit, Mohan, and the third person, sitting between them, as a single personality, we can arrange them in 17! ways but Mohit and Mohan can also be arranged in 2 ways.
Required number of permutations = 18 \times (17!) \times 2 = 2 \times 18!

Correct answer is 72
Explanation:
All the boxes contain a distinct number of chocolates.
For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on.
The number of ways of giving 4 boxes to the 4 person is: _{4}^{8}\textrm{C}= 70

Correct answer is 65000
Explanation :
Out of 26 alphabets two distinct letters can be chosen in 26P_{2} ways. Coming to numbers, there are 10 ways.(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10 ways to choose the second digit. Hence there are totally 10 \times 10 = 100 ways.
Combined with letters there are 6P_{2} \times 100 ways = 65000 ways to choose vehicle numbers.

the correct answer is 252
Explanation:
Since there are 5 cups of each kind, prepared with milk or tea leaves added first, are identical hence, the total number of different ways of presenting the cups to the expert is \frac{10!}({5! \times 5!}= 252 [/latex

Correct answer is 27404
Explanation:
Given total 16 Red roses and 14 White roses = 30 roses
Four flowers have to be selected from 30 i.e, _{4}^{30}\textrm{C}= 27405 Ways
Now, at least one Red rose is selected i.e, 27405(total) - 1(all four are white roses) = 27404 ways.

Correct answer is 30
Explanation:
Since order does not matter it is a combination. The word AND means multiply.
Given 4 basketball, 3 volleyball, 2 soccer.
We want 2 basketball games and 1 other event. There are 5 choices left.
C(n,r)
C(How many do you have, How many do you want)
C(have 4 basketball, want 2 basketball) x C(have 5 choices left, want 1)
C(4,2) \times C(5,1) = (6)(5) = 30
Therefore there are 30 different ways in which you can go to two basketball games and one of the other events