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Permutation and Combination Questions

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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

Answer: Option D

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

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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

Answer: Option A

Explanation:

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

We have to distribute three 2's to a, b, c in 3+3−1C3−1 = 5C2 = 10 ways

We have to distribute four 3's to a, b, c in 3+4−1C3−1 = 6C2 = 15 ways

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

Total ways = 10*15*3 = 450 ways

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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

Answer: Option E

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!.

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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

Answer: Option A

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

Answer: Option B

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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A lady has fine gloves and hats in her closet- 14 blue, 20 red, and 18 yellow. The lights are out and it is totally dark. In spite of the darkness, she can make out the difference between a hat and a glove. She takes out an item out of the closet only if she is sure that if it is a glove. How many gloves must she take out to make sure she has a pair of each color?


A38

B40

C34

DNone of these

Answer: Option B

Explanation:

Additon of 2 max. values i.e (20+18)
then add 2 to that value to get the pair for smallest color gloves
=> (20+18)+2=40

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How many different integers can be expressed as the sum of three distinct numbers from the set {3, 10, 17, 24, 31, 38, 45, 52}?


A8

B56

C16

D15

Answer: Option C

Explanation:

In this question we are not asked how many ways we can select 3 numbers out of 8.
But how many different numbers can be expressed as a sum of three numbers from the given set.
For example, 3 + 10 + 31 = 3 + 17 + 24 = 47. So 47 can be expressed as a sum of 3 numbers in two different ways but 47 should be considered as only one number.
Now the minimum number that can be expressed as a sum of 3 numbers = 30. (3+10+17)
The next number is 37.
Similarly the largest number is 38 + 45 + 52 = 135.
So there exists many numbers in between, with common difference of 7.
Total numbers = (l?a)/d +1 = (135-30)/7 +1= 15+1=16.

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A college has 10 basketball players. A 5 member team and a captain will be selected out of these 10 players. How many different selections can be made?


A1200

B126

C1260

D1350

Answer: Option C

Explanation:

We can select the 5 member team out of the 10 in 10C5 ways = 252 ways. The captain can be selected from amongst the remaining 5 players in 5 ways. Therefore, total ways the selection of 5 players and a captain can be made = 252*5 = 1260.

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In how many ways can 5 different toys be packed in 3 identical boxes such that no. box is empty, if any of the boxes may hold all of the toys?


A25

B125

C15

D243

Answer: Option A

Explanation:

Number of ways of achieving the second option 3 - 1 - 1

Three toys out of the 5 can be selected in 5C3ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.

Therefore, total number of ways of getting the 3 - 1 - 1 option is 5C3 = 10 = 10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes

= number of ways of achieving Case a + number of ways of achieving Case b
= 15 + 10 = 25 ways.

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How many 4-bit digit numbers that do not contain the digits 3 or 6 are there?


A5040

B4096

C7200

D3584

Answer: Option D

Explanation:

Let the number be _ _ _ _

Left digit can be any of the (1,2,4,5,7,8,9) --> 7 ways
Other digits can be (0,1,2,4,5,7,8,9) --> 8 ways

So the total ways = 7*8*8*8 = 3584 or 3584 numbers without any 3 or 6.

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