AMCAT Permutation and Combinations Quiz-1

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 EA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with EA.
The next word starts with EG and then A, i.e., EGA _ _.
The first of the words will be EGART. The next word will be EGART.
Therefore, the rank of EGART will be 24+6+2= 32

There are a total 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

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

Let last digit is 2 ,
When the second last digit is 4 the remaining 4 digits can be filled in 120 ways, similarly second last digit is 6, 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.

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!

To form a rectangle. we require two lines from East to West and two lines from North to South. Therefore, total number at rectangles which we can get is:
^{p}\textrm{C}_{2} \times ^{q}\textrm{C}_{2} = \frac{p(p-1)}{2} \times \frac{q(q-1)}{2} = \frac{(pq)(p-1)(q-1)}{4}

Initially there were ^{12}\textrm{C}_{2} = 66 lines. Now for the four ants among themselves the number of lines are decreased by[/latex] (^{4}\textrm{C}_{2} - 1) = 5 [/latex] i.e., a net decrease of 5 lines i:e There are 66 - 5 = 61 possible lines.

No. of ways in which at least one yellow ball can be selected = 2^{5} - 1 = 31
No. of ways in which at least one black ball can be selected = 2^{4}-1 = 15
No. of ways in which one or more red balls can be selected = 2^{3} = 8,
Therefore, total number of ways = 31 \times 15 \times 8 = 3720

Let the spider's legs be numbered 1. 2, 3, …, 8
Let a_{k}, and b_{k} , denote the Sock and the shoe respectively for the K th leg where 1 ≤ k ≤ 8.
So a total of sixteen objects are there which are to be arranged such that a_{k} comes before b_{k}
16 objects can be arranged in 16! ways, Out of these the arrangements in which a_{k} comes before b_{k} = 12 \times 16!
Similarly out of the remaining, the arrangement in which a_{2} comes before b_{2} =12 \times 12 \times 16!
Finally, arrangements in which a_{k} comes before b_{k}= (\frac{1}{2})^{8} \times 16!

We know that, the number of straight lines that can be formed by the 11 flags in which 6 flags are collinear except those that can be selected out of these 6 flags are collinear.
Hence, the required number of straight lines
= ^{11}\textrm{C}_{2} - ^{6}\textrm{C}_{2} - ^{5}\textrm{C}_{2} + 1 + 1
= 55 - 15 - 10 + 2 = 32