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?
Answer:
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?
Answer:
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??
Answer:
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
=2C1*5C4*9C6=840
Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players
=2C1*5C5*9C5=252
=> 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



6 comments:

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

    ReplyDelete
    Replies
    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.
      We need your help too, please do support us by sharing this blog with all your friends on Facebook.

      Cheers!

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