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# Test: 35 Year JEE Previous Year Questions: Permutations and Combinations

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## 22 Questions MCQ Test Mathematics For JEE | Test: 35 Year JEE Previous Year Questions: Permutations and Combinations

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Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 1

### Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are      [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 1

Required number of numbers = 5 x 6 x 6 x 4 = 36 x 20 = 720.

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 2

### Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 2

Required number of numbers = 3 x 5 x 5 x 5 = 375

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 3

### Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are [2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 3

We know that a number is divisible by 3 only when the sum of  the digits is divisible by 3. The given digits are 0, 1, 2, 3, 4, 5.
Here the possible number of combinations  of 5 digits out of 6 are 5C4 = 5, which are as follows
– 1 + 2 + 3 + 4 + 5 = 15 = 3 × 5
0 + 2 + 3 + 4 + 5 = 14 (not divisible by 3)
0 + 1 + 3 + 4 + 5 = 13 (not divisible by 3)
0 + 1 + 2 + 4 + 5 = 12 =  3 × 4
0 + 1 + 2 + 3 + 5 = 11 (not divisible by 3)
0 + 1 + 2 + 3 + 4 = 10  ( not divisible by 3)
Thus the number should contain the digits 1, 2, 3, 4, 5 or  the digits 0, 1, 2, 4, 5.
Taking 1, 2, 3, 4, 5, the 5 digit numbers are = 5! = 120
Taking 0, 1, 2, 4, 5, the 5 digit numbers are = 5! – 4! = 96
∴  Total number of numbers = 120 + 96 = 216

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 4

The sum of integers from 1 to 100 that are divisible by 2 or 5 is[2002]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 4

Required sum = (2 + 4 + 6 + ... + 100) + (5 + 10 + 15 + ... + 100)
– (10 + 20 + ... + 100)
= 2550 + 1050 – 530 = 3050.

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 5

If nCr denotes the number of combination of n things taken r at a time, then the expression nCr +1 + nC r -1 + 2 x nCr equals[2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 5

n Cr +1 + n Cr -1+ 2nCr = nCr -1+ n Cr + nCr nCr +1

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 6

A student is to answer 10 out of  13   questions  in  an examination such that he must choose at least 4 from  the first five questions. The number of choices available to him is[2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 6

As for given question two cases are possible.
(i) Selecting  4 out of first five question and 6  out  of remaining 8 question
= 5C4 x 8 C6 = 140 choices.
(ii) Selecting 5 out of first five question and 5  out of remaining  8 questions
= 5C5 x 8 C5 = 56 choices.
∴ total number of choices =140 + 56 = 196.

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 7

The number of ways in which 6 men and 5 women can dine at a round table  if no two women   are to sit together is given by[2003]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 7

No. of ways in which 6 men can be arranged at a round table = (6 - 1)! = 5!
Now women can be arranged in 6 P5 = 6! Ways.
Total Number  of ways  = 6! × 5!

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 8

How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 8

Total number of arran gemen ts of letters in the word GARDEN = 6 ! = 720 there are two vowels A and E, in half of the arrangements A preceeds E and other half A follows E.
So, vowels in alphabetical order in x 720 = 360

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 9

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is [2004]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 9

We know that the number of ways of distributing n identical items among r persons, when each one of them receives at least one item is n -1Cr -1
∴ The required number of ways

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 10

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number [2005]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 10

Alphabetical order is A, C, H, I, N, S
No. of words starting with A – 5!
No. of words starting with C – 5!
No. of words starting with H – 5!
No. of words starting with I – 5!
No. of words starting with N – 5!
SACHIN – 1
∴ sachin appears at serial no 601

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 11

At an election , a voter may vote for an y number  of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least  one candidate, then the number of ways in which he can vote is [2006]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 11

10C1 + 10C2 + 10C3 + 10C4
= 10 + 45 + 120 + 210 = 385

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 12

The set S  = {1, 2, 3, ......., 12} is to be partitioned into three sets A , B, C of equal size. Thus A ∪ B ∪ C = S , A∩B =B∩C= A∩C= φ. The number of ways to partition S is [2007]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 12

Set S = {1, 2, 3, ...... 12}

A∪B∪C = S, A∩B =B∩C=A∩C= φ.

∴ The number of ways to partition

= 12C4 × 8C4 × 4C4

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 13

In a shop there are five types of ice-creams available. A child buys six ice-creams.Statement-1 : The number of different ways the child can buy the six ice-creams is 10C5.Statement -2 : The number of different ways the child can buy the six ice-creams is equal to the number of different ways of arranging 6 A’s and 4 B’s in a row. [2008]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 13

The given situation in statement 1 is equivalent to find the non negative integral solutions of the equation x1 + x2 + x3 + x4 + x5 = 6
which is coeff. of x6 in the expansion of (1 + x + x2 + x3 + .....∞)5
= coeff. of x6 in (1– x)–5
= coeff. of x6 in 1 + 5x +

∴ Statement 1 is wrong.
Number of ways of arranging 6A’s and 4B’s in a row

which is same as the number of ways the child can buy six icecreams.
∴ Statement 2 is true.

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 14

How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent?

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 14

First let us arrange M, I, I, I, I, P, P Which can be done in

ways
√ M √ I √ I √ I √ I √ P √ P √

Now 4 S can be kept at any of the ticked places in 8C4 ways so that no two S are adjacent.
Total required ways

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 15

From 6 different novels and 3 different dictionaries,4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. Then the number of such arrangement is: [2009]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 15

4 novels, out of 6 novels and 1 dictionary out of 3 can be selected in 6 C4 x 3C1 ways
Then 4 novels with one dictionary in the middle can be arranged in 4! ways.
∴ Total ways of arrangement = 6C4 x 3C1x 4! = 1080

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 16

There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is [2010]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 16

Total number of ways = 3C2 x 9C2

= 3 x 36 = 108

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 17

Statement-1: The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box is empty is 9C3 .Statement-2: The number of ways of choosing any 3 places from 9 different places is  9C3 . [2011]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 17

The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box empty is same as the number of ways of selecting (r – 1) places out of (n – 1) different places, that is n-1Cr -1 .
Hence required number of ways = 10-1C4-1 = 9C3
∴  Both statements are correct and second statement is a correct explanation of statement -1.

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 18

These are 10 points in a plane, out of these 6 are collinear, if N is the number of triangles formed by joining these points. then:[2012]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 18

Number of required triangles

=  10C36C3

= 120 - 20 =100

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 19

Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is : [2012]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 19

Number of white balls = 10
Number of green balls = 9

an d Number of black balls = 7
∴ Required probability = (10 + 1) (9 + 1) ( 7 + 1)  – 1 = 11.10.8 –1 = 879

[∵ The total number of ways of selecting one or more items from p identical items of one kind, q identical items of second kind; r identical items of third kind is ( p + 1) (q + 1) (r + 1) –1 ]

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 20

Let Tn be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1 – Tn = 10, then the value of n is : [JEE M 2013]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 20

We know, Tn = nC3, Tn+1 = n+1C3
ATQ, Tn+1 – Tn = n+1C3nC3 = 10
nC2 = 10
⇒ n = 5.

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 21

The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is : [JEE M 2015]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 21

Four digits number can be arranged in 3 × 4! ways.
Five digits number can be arranged in 5! ways.
Number of integers = 3 × 4! + 5! = 192.

Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 22

If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is : [JEE M 2016]

Detailed Solution for Test: 35 Year JEE Previous Year Questions: Permutations and Combinations - Question 22

ALLMS
No. of words starting with

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