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Permutation and Combination Questions

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21 / 100

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When 3 dice are simultaneously thrown, which of the following would be obtained?

AResult 1: A 5, a 3 and a 6

BResult 2: A 5 three times

CResult 3: A 5 twice and a 3

DAll three results are equally likely.

Explanation:

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30
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22 / 100

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In a cycle race there are 5 persons named as J,K,L,M,N participated for 5 positions. How many number of ways can M finishes always before N?

A55

B50

C60

D36

Explanation:

No. of ways in which 5 persons can finish is 5! = 120 (there are no ties)
Now, in half of these ways M can finish before N.

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23 / 100

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There are 6561 balls out of them 1 is heavy. Find the min. no. of times the balls have to be weighed for finding out the heavy ball.

A12

B8

C18

D6

Explanation:

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24
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6 M:30 S
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24 / 100

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In a dance school, there are 14 girls and 15 boys. In how many ways, a group consisting of 3 girls and 7 boys can be formed for a competition?

A2342340

B2232340

C2132840

D2132340

Explanation:

3 girls selected from 14 girls in 14C3 ways = 14! / 3! (14 - 3)! = 14! / 3!11! = 364 ways. (Formula used nCr = n! / r! (n-r)!)
And 7 boys selected from 15 boys in 15C7 ways = 15! / 7! (15 - 7)! = 15! / 7!8! = 6435 ways
So, possible ways = 364 x 6435 = 2342340 ways.

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25
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25 / 100

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The players for district football team are selected from three different colleges and each college has 5 top players. Except captain and goalkeeper, each 3 are selected from each 3 colleges. Find the number of possible ways of such selection.

A1500

B1000

C4500

DNone of these

Explanation:

Given, total number of players = 11
2 players(captain & goalkeeper)selected in 1 way.
Now, use the combination method, nCr = n! / r! (n-r)!
3 out of 5 can be selected in 5C3 ways = 5! / 3! (5 - 3)! = 10
Similarly, other 3 are selected from other two colleges.
Hence, total number of possible ways = 1 * 10 * 10 * 10 = 1000

Workspace

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24
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Solv. In. Corr.
80
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In a college cricket club of 22 players, captain and wicketkeeper are selected from seniors team. For remaining players they have to select from 8 seniors and 12 juniors. Now, for a new team of 11 players except captain and wicketkeeper, out of 9 they select 4 from juniors and 5 from seniors. Find the number of methods to select players of new team.

A12220

B27620

C15870

D27720

Explanation:

New team of 11 players will be formed.
As per quesiton 5 seniors selected out of 8 in 8C5 ways = 8! / 5! (8 - 5)! = 8! / 5!3! = 56 ways
and 4 juniors selected out of 12 in 12C4 ways = 12! / 4! (12 - 4)! = 495 ways
nCr = n! / r!(n-r)! <=== point to remember
2 players [captain & wicketkeeper] already selected = 1 possible way.
Hence the total number of possible ways = 56 x 495 x 1 = 27720 ways.

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If there are 30 cans out of them one is poisoned if a person tastes very little he will die within 14 hours so if there are mice to test and 24 hours to test, how many minimum mices are required to find the poisoned can?

A3

B2

C6

D1

Explanation:

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Six friends decide to share a big cake. Since all of them like the cake, they begin quarreling who gets to first cut and have a piece of the cake. One friend suggests that they have a blindfold friend choose from well shuffled set of cards numbered one to six. You check and find that this method works as it should simulating a fair throw of a die. You check by performing multiple simultaneous trials of picking the cards blindfold and throwing a die. You note that the number shown by the method of picking up a card and throwing a real world die, sums to a number between 2 and 12. Which total would be likely to appear more often- 8,9 or 10?

A8

BAll are equally likely

C9

D10

Explanation:

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A white cube(with six faces) is painted red on two different faces. How many different ways can this be done (two paintings are considered same if on a suitable rotation of the cube one painting can be carried to the other)?

A2

B15

C4

D30

Explanation:

Paint on 1 side of the cube.

The condition given here is when on suitable rotation the paint is carried to another which means when a face is turned the adjacent side is having the same paint. So all the 5 sides will be adjacent to one sideâ€¦

One will be exactly opposite to that sideâ€¦

Selecting 1 adjacent side = 1 way

Selecting the opposite one=  1 way  So total=  2 ways.

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From 5 men and 11 women, in how many ways can a panel of 11 be formed such that the number of men is not more than 3?

A1650

B2255

C5522

DNone of these

Explanation:

5C0*11C11+5C1*11C10+5C2*11C9+5C3*11C8=2256

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