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Permutation and Combination Questions

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There are 16 teams divided in 4 groups. Every team from each group will play with each other once. The top 2 teams will go to the next round and so on the top two teams will play the final match. Minimum how many matches will be played in that tournament?


A43

B40

C14

D50

Answer: Option A

Explanation:

: In each group, total matches played = 4C2 = 6.
So total matches played in the first round = 6 * 4 = 24

Now top two teams from each group progress to the next round.
Now these 8 teams are pooled into 2 groups of 4 teams
. Total matches played in the second round = 6 * 2 = 12. (4C2=6 in each grp)

Now 4 teams progress to the next round.
Total matches played in the third round = 4C2 =6

From this round, 2 teams progress to the next round. And final will be played between them.
Total matches = 24 + 12 + 6 + 1 = 43

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Oranges can be packed in sets of 10 oranges in box type A or 25 oranges in box type B. A carton comprising of 1000 oranges of type a and b is packed. How many different combinations are possible in the number of type A and type B boxes while organizing the oranges?


A21

B20

C19

D18

Answer: Option A

Explanation:

The expression will be 10x+25y=1000
the above exp will be satisfied only with multiples of 2 up to 38.
so no. of combinations are 19.
if oranges are packed in type A boxes only= 1 way
if oranges are packed in type B boxes only= 1 way
so total = 19+1+1=21

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A number plate can be formed with two alphabets followed by two digits, with no repetition. Then how many possible combinations can we get?


A58500

B67600

C65000

D64320

Answer: Option A

Explanation:

26 * 25 * 10 * 9 = 58500

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All even numbers from 2 to 98 inclusive, except those ending 0, are multiplied together. What is the rightmost digit (the units digit) of the product?


A6

B2

C0

D4

Answer: Option A

Explanation:

2^10 * 4^10 * 6^10*8^10 = 2^70 * 3^10
2^2 * 3^2 = 36
So, the right most digit = 6

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There is a 7-digit telephone number with all different digits. If the digit at extreme right and extreme left are 5 and 6 respectively, find how many such telephone numbers are possible?


A120

B30240

C6720

DNone of these

Answer: Option C

Explanation:

If the digit at extreme right and extreme left are 5 and 6 respectively.
So, the 7 digit number is : 6 _ _ _ _ _ 5
Since repetition of digits are not allowed the second digit can be chosen from (0,1,2,3,4,7,8,9) 8 of the remaining digits.
5 spaces are left, hence 8P5=6720

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In how may ways can 'mn' things be distributed equally among n groups


AmnPm * mnPn

BmnCm * mnCn

CmnCn

D(mn)! / (m!)^n * n!

Answer: Option D

Explanation:

mn into n groups
= (mn)! / (m! m! ....n times * n!(n groups are similar))
= (mn)! / (m!)^n * n!

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There are 20 persons sitting in a circle. In that there are 18 men and 2 sisters. How many arrangements are possible in which the two sisters are always separated by a man?


A18!x2

B17!

C17x2!

D12

Answer: Option A

Explanation:

Let the first sister name is A. Now she can sit any where in the 20 places (Symmetrical). Now her sister B can sit to her left or right in 2 ways. Now the remaining 18 persons can be sit in 18 places in 18! ways. Total = 18! * 2

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In how many ways a team of 11 must be selected a team 5 men and 11 women such that the team must comprise of not more than 3 men.


A1565

B2256

C2456

D1243

Answer: Option B

Explanation:

5 men and 11 women and not more than 3 men
1st way= 1men and 10 women= 5C1 * 11C10
2nd way= 2 men and 9 women=5C2 *11C9
3rd way= 3 men and 8 women= 5C3* 11C8
4th way= 0 men and 11 women=5C0* 11C11
Adding all we get= 2256.

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There are ten points marked on a straight line, and another 11 points marked on a line parallel to this one. How many triangles could be formed using these points ?


A495

B550

C1045

D2475

Answer: Option C

Explanation:

10C2 * 11C1 + 10C1 * 11C2

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A manufacturer of chocolates makes 6 different flavors of chocolates. The chocolates are sold in boxes of 10.
How many "different" boxes of chocolates can be made?


A2003

B3003

C1231

D3013

Answer: Option B

Explanation:

If n similar articles are to be distributed to r persons, x1+x2+x3......xr=n each person is eligible to take any number of articles then the total ways are n+r?1Cr?1
In this case x1+x2+x3......x6=10

In such a case the formula for non negative integral solutions is n+r?1Cr?1 Here n = 6 and r = 10. So total ways are 10+6?1C6?1 = 3003

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