CAT Free Online Test - Permutation Answers

Below are detailed answers for each of the questions asked in test, the explanations given below would further strengthen your understanding of Permutation & Combinations topic. 

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Q1. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.

Q2.In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
Required number of ways           = (8C5 x 10C6)
= (8C3 x 10C4)
=           8 x 7 x 6           x           10 x 9 x 8 x 7   
3 x 2 x 1            4 x 3 x 2 x 1
= 11760.

Q3.In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together??
Let 4 girls be one unit and now there are 6 units in all.
They can be arranged in 6! ways.
In each of these arrangements 4 girls can be arranged in 4! ways.
=> Total number of arrangements in which girls are always together
=6!*4!=720*24= 17280

Q4. 12 points lie on a circle. How many cyclic quadrilaterals can be drawn by using these points?
For any set of 4 points we get a cyclic quadrilateral. Number of ways of choosing 4 points out of 12 points is 12C4=495
. Therefore, we can draw 495 quadrilaterals.​

Q5. The Indian Cricket team consists of 16 players. It includes 2 wicket keepers and 5 bowlers. In how many ways can a cricket eleven be selected if we have to select 1 wicket keeper and at least 4 bowlers?
We are to choose 11 players including 1 wicket keeper and 4 bowlers
or, 1 wicket keeper and 5 bowlers.
Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players
Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players
=> Total number of ways of selecting the team:
=840+252= 1092

Q6. How many factors of (2^4)×(5^3)×(7^4) are odd numbers?
Any factor of this number should be of the form 2a×3b×5c.
For the factor to be an odd number, a should be 0.
b can take values 0,1, 2,3, and c can take values 0, 1, 2,3, 4.
Total number of odd factors =4×5= 20

Q7. How many factors of (2^5)×(3^6)×(5^2) are perfect squares?
Any factor of this number should be of the form 2a×3b×5c.
For the factor to be a perfect square a, b, c have to be even.
a can take values 0, 2, 4. b can take values 0,2, 4, 6 and c can take values 0,2.
Total number of perfect squares =3×4×2= 24

Q8. In how many ways can eight directors, the vice-chairman and chairman of a firm be seated at a round table, if the chairman has to sit between the vice-chairman and the director?
In a circular arrangement problem, we always fix one position and the look at ways of arranging the rest with respect to the fixed position.

Case 1 - seating of chairman - In this question, we fix the position of the Chairman. Thus the Chairman can be seated in 1 way only.

Case 2 - seating of vice-chairman - The vice-chairman can be to the left or right of the chariman. Thus the vice chairman can be seated in 2 ways.

Case 3 - seating of the 8 directors - The rest of the 8 directors can be seated in 8 positions in 8! ways. This is because the first director can be seated in andy of 8 positions in 8 ways. And then the second director can be seated in any of the remaning 7 positions in 7 ways and so on, thus giving the total number of ways of seating the 8 director as 8 * 7 * 6* 5 * 4 * 3 * 2 * 1, i.e. 8! ways.

Thus the desired answer is when case 1 and 2 and 3 happen together, which is 1 * 2 * 8! = 2 * 8! ways.

Q9. In how many ways can 15 people be seated around two round tables with seating capacities of 7 and 8 people?
‘n' objects can be arranged around a circle in (n - 1)!.

If arranging these 'n' objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number.
i.e., number of arrangements = (n-1)!/2.
You can choose the 7 people to sit in the first table in 15C7 ways.
After selecting 7 people for the table that can seat 7 people, they can be seated in (7-1)! = 6!.
The remaining 8 people can be made to sit around the second circular table in (8-1)! = 7! Ways.

Hence, total number of ways 15C8 * 6! * 7!

Q10. Suppose you can travel from a place A to a place B by 3 buses, from place B to place C by 4 buses, from place C to place D by 2 buses and from place D to place E by 3 buses. In how many ways can you travel from A to E?
The bus from A to B can be selected in 3 ways.
The bus from B to C can be selected in 4 ways.
The bus from C to D can be selected in 2 ways.
The bus from D to E can be selected in 3 ways.
So, by the General Counting Principle, one can travel from A to E in 3*4*2*3= 72 ways

Q11. How many words can be formed by using 4 letters at a time of word "SURPRISE" (words may be meaningful or meaningless according to dictionary)
(1) No. of words having all letters different = 6*5*4*3 = 360
(2) No. of words having any one letter repeated
=(2*1)*(5*4*3*2*1/3*2*1*2*1)*(4*3*2*1/2*1) = 240
(3) No. of words having two letter repeated = (1)*(4*3*2*1/2*1*2*1) = 6
Total words = 360+240+6 = 606


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  2. cant understand Q11 solution!! can anyone pls explain

    1. Hi PalaneSami,

      Below is the detailed answer:
      List & count of alphabets which are part of word SURPRISE:
      E=1, I=1, P=1, R=2, S=2 & U=1
      which is 6 unique letters (EIPRSU - Lets represent any letter of this unique set as X) and 2 repeated letters(R & S - lets represent any letter of this repeated set as Y).
      We are asked to form different 4 letter words using words in surprise. Looking at the composition of letter of the word surprise this could be done as follows:
      1. XXXX - Using all unique letters i.e. all 4 letters from EIPRSU
      No. of words having all letters different = 6*5*4*3 = 360
      2. XXXY - Using 3 unique and 1 repeated letter i.e.
      2 letters from EIPRU and 2 letters of S OR
      2 letters from EIPSU and 2 letters of R
      Hence No. of words having any one letter repeated
      =(2*1)*(5*4*3*2*1/3*2*1*2*1)*(4*3*2*1/2*1) = 240
      3. XXYY - Using all letters from repeated letters set i.e. all 4 from RRSS set
      Hence No. of words having two letter repeated = (1)*(4*3*2*1/2*1*2*1) = 6
      Total words = 360+240+6 = 606

      Hope this gives you better understanding, let us know if you have any doubts.
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