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# Permutation and Combination Questions

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82
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41 / 100

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how many matches would have to be played to decide a winner in a knockout tournament consisting of eight teams?

A7

B6

C8

D5

Explanation:

1 match for 2 team, first 8 team = 4 match
4 team - 2 match
2 team - 1 match
So, Total = 4+2+1=7 match

Workspace

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SHSTTON
55
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15
Solv. In. Corr.
70
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Five different roads join a village to nearby city. The no. of different ways in which a person can go to town and come back is ?

A25 ways

B35 ways

C22 ways

D45 ways

Explanation:

If we choose 1st road, To come back we have 5 ways..
so total ways = 5+5+5+5+5 = 25 ways

Workspace

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SHSTTON
19
Solv. Corr.
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Solv. In. Corr.
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43 / 100

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The number of ways in which 15 students A1, A2 -------A15 can be ranked, such that A4 is always above A8 is

A15!

B13!

C15!/2

D13!/2

Explanation:

When A8 is in 2nd position A4 can occupy only 1st.
When A8 is in 3rd position A4 can occupy only 1st or 2nd.
So the number of ways in which A4 and A8 can be arranged = 1+2+...+14 = 14*15/2
In the remaining 13 positions other students can be permuted in 13! ways.
Hence total number of ways = 13! * 14*15/2 = 15!/2

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In how many ways can the letters of the word "BANKING" be arranged, such that vowels do not come together?

A1800

B1600

C1890

DNone of these

Explanation:

Total no. of words = 7!/2!=2520
Total no. of words in which vowels together = bnkng(ai)
=(6!/2!)*2!=720

So, the number of words in which vowels do not come together = (2520 - 720) = 1800

Workspace

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SHSTTON
8
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19
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27
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45 / 100

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A class photograph has to be taken. The front row consists of 6 girls who are sitting. 20 boys are standing behin The two corner positions are reserved for the 2 tallest boys. In how many ways can the students be arranged?

A6! * 1440

B18! * 1440

C18! *2! * 1440

DNone of these

Explanation:

Two tallest boys can be arranged in 2! ways, rest 18 can be arranged in 18! ways.
Girls can be arranged in 6! ways.
Total number of ways in which all the students can be arranged = 2! * 18! * 6! = 18! *1440

Workspace

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SHSTTON
11
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32
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The letter of the word LABOUR are permuted in all possible ways and the words thus formed are arranged as in a dictionary. What is the rank of the word LABOUR?

A275

B251

C240

D242

Explanation:

The order of each letter in the dictionary is ABLORU.
Now, with A in the beginning, the remaining letters can be permuted in 5! ways.
Similarly, with B in the beginning, the remaining letters can be permuted in 5! ways.
With L in the beginning, the first word will be LABORU, the second will be LABOUR.
Hence, the rank of the word LABOUR is 5!+5!+2 =242

Workspace

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SHSTTON
7
Solv. Corr.
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24
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47 / 100

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In how many possible ways can write 3240 as a product of 3 positive integers a,b and c.

A450

B420

C350

D320

Explanation:

3240 = 23*34*51 = a*b*c

We have to distribute three 2's to a, b, c in 3+3âˆ’1C3âˆ’1 = 5C2 = 10 ways

We have to distribute four 3's to a, b, c in 3+4âˆ’1C3âˆ’1 = 6C2 = 15 ways

We have to distribute one 5 to a, b, c in 3 ways.

Total ways = 10*15*3 = 450 ways

Workspace

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9
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23
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32
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In how many ways can the letters of the english alphabet be arranged so that there are seven letter between the letters A and B, and no letter is repeated

A24P7 * 2 * 18!

B36 * 24!

C24P7 * 2 * 20!

D18 * 24!

EBoth A & B

Explanation:

We can fix A and B in two ways with 7 letters in between them. Now 7 letters can be selected and arranged in between A and B in 24P7 ways. Now Consider these 9 letters as a string. So now we have 26 - 9 + 1 = 18 letters
These 18 letters are arranged in 18! ways.
So Answer is 2 x 24P7 x 18!
OR 2 x 24P7 x 18! = 36 x 24!.

Workspace

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How many different integers can be expressed as the sum of three distinct numbers from the set {3, 8, 13, 18, 23, 28, 33, 38, 43, 48}?

A22

B8

C56

DNone of these

Explanation:

minimum possible number = 24 ( 3+8+13)

maximum possible number = 129 ( 38+43+48)

so if we assume a AP series now where common difference is 5 and the series will start with 24 and last term will be 129 then, using AP formula

Total numbers = [(l-a)/d + 1] = [(129 - 24)/7+1 ] = 22

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A owes B Rs.50. He agrees to pay B over a number of consecutive days starting on a Monday, paying single note of Rs.10 or Rs.20 on each day. In how many different ways can A repay B.

A10

B8

C18

D16

Explanation:

He can pay by all 10 rupee notes in 5 days = 1 way
3 Ten rupee + 1 twenty rupee = 4!/(3!*1!) = 4 ways
1 Ten rupee + 2 twenty rupee notes = 3!/(2!*1!) = 3 ways
Total ways = 1 + 4 + 3 = 8

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