JEE Advanced Level Test: Permutation and Combinations- 1 - JEE MCQ

Selecting 3 consonants out of 7,
No. of ways = ⁷C₃ 
Selecting 2 vowels out of 4,
No. of ways = ⁴C₂ 
No. of ways of forming the words
= ⁷C₃* ⁴C₂* 5! (for arrangement of the letters)
= 35 * 6 * 120
= 25200 

Given, 
Number of visitors= 4
Number of hotels= 5
So, A can disperse among 5 hotels,
      B can disperse among 4 hotels,
      C can disperse among 3 hotels,
      D can disperse among 2 hotels.
∴ They can disperse in the hotels in 5 × 4 × 3 × 2 = 120 ways

Number of words beginning with A = 4! = 4 × 3 × 2 = 24
Number of words beginning with N = 4! = 4 × 3 × 2 = 24
Number of words beginning with R = 4! = 4 × 3 × 2 = 24
Number of words beginning with U = 4! = 4 × 3 × 2 = 24
So, in total 96 words will be formed while beginning with letter A, N, R and U.
Now, 97th word according to dictionary order will be VANRU
Now, 98th word according to dictionary order will be VANUR
Now, 99th word according to dictionary order will be VARNU
Finally, 100th word according to dictionary order will be VARUN.
Hence, option (c) is the correct answer.

Total number of jackets=3
Total number of shirts=10
Total number of pair of slacks=5
Ways of selecting one jacket =3C1=3
Ways of selecting one shirt =10C1=10
Ways of selecting one pair of slack = 5C1=5
Number of different outfits that can be made=3×10×5=150
Number of different outfits that can be made=150

i) It's 6*4 because there are 6 different roads to get to B, then after getting there, can take 4 different paths. So every one road to B has 4 different possible paths. 

i) There are 5 places to fill

2 places of alphabet(out of 26) can be filled in =26∗25 ways

3 places of digits(out of 10) can be filled in = 10∗9∗8 ways

No of different plates = 26∗25∗10∗9∗8=468000

ii) We have total four place to cover with two alphabets and two digits.

For the first place:

Out of 26 alphabets we need one  

∴ combinations for first place = ²⁶C₁ = 26

For the second place:

We are left with 25 alphabets now as no repetitions are allowed.

∴combinations for second place= ²⁵C₁ = 25

For the third place:

Out of 10 digits we need to select one

∴combinations for third place = ¹⁰C₁ = 10

For the fourth place:

We are left with 9 digits now as no repetitions are allowed

∴combinations for fourth place= ⁹C₁ = 9

∴Total Combinations of number plates = 26 * 25 * 10 * 9 = 58,500

i) 6!/(6-3)!=6*5*4=120

.

ii) 2/6 * 120 = 40

.

iii) 2/6 * 120 = 40

.

iv) 1/6 * 120 = 20

Different number of ways = 3⁵
= 243

A 3-digit no.: _ _ _
Possible digits: 0, 1 ,2, 3, 4, 5, 6, 7, 8, 9
In the first place there are 9 possible digits (0 cannot be the 1st digit)
In the second and third place there are 10 possible digits.
Total no. of 3-digit numbers = 9 *10 *10 = 900 
For 3-digit no. without digit 7,
Possible digits: 0, 1 ,2, 3, 4, 5, 6, 8, 9
In the first place there are 8 possible digits (0 cannot be the 1st digit)
In the second and third place there are 9 possible digits.
Total no. of 3-digit numbers = 8 *9 *9 = 648 
No. of 3-digit numbers with at least one digit 7 = 900 – 648 = 252

For 5 lettered palindrome, we can fill the first 3 letters randomly and then the last 2 letters will become fixed.

there are 26 letters.
therefore the number of ways =26×26×26=26³
 = 17,576.

We are drawing two cards without replacement from a deck of 52 cards.

No. of ways = ⁵²C₁ x ⁵¹C₁ = 52 x 51 = 2652

Consider the words starting with A.
A−−−−
Number of ways of filling the 2nd place is 4
Number of ways of filling the 3rd place is 3
Number of ways of filling the 4th place is 2
Number of ways of filling the 5th place is 1
Hence in total 4×3×2×1=24
Similar cases for the words starting with I.=24
Hence 2×24 =48
Consider the words ending with A−−−−A
Number of ways of filling the 1st place is 4
Number of ways of filling the 2nd place is 3
Number of ways of filling the 3rd place is 2
Number of ways of filling the 4th place is 1
Hence in total 24 ways.
Similarly for I- 24 ways.
Hence total number of words 4×24=96
But this includes the words which start and end with Vowels.
Hence A−−−I
Number of such words is 3!.
Similarly I−−−A.
Number of such words is 3!.
Hence total number of words beginning with or ending with Vowels is
= 96−3!−3!=96−12
= 84

The odd digits which can be considered are (1,3,5,7 and 9).
If we start from left, the first digit can be any number out of the five digits. The second place can be filled with any four digits except the digit at the first place.
Similarly, four ways for the third digit until we reach the tenth place.
Therefore, the total numbers formed are 5 × 49

There can 5 girls selected for the first seat.

Then, there are 4 girls to select from for the second seat.

Then, there are 3 girls to select from for the third seat.

Then, there are just 2 girls to select from for the fourth seat.

Then, there is just 1 girl for the fifth and final seat.

Therefore they number of ways to seat 5 girls in 5 seats is:

5×4×3×2×1⇒20×6×1⇒120×1⇒120

So there are 120 different ways to seat 5 girls in 5 chairs.

Total 3 digit numbers = 900
There are only 5 odd numbers (1,3,5,7,9)
So 3 digit numbers having containing only odd digits = 5*5*5 = 125
So, 3 digit numbers containing at least one even digit = 900–125 = 775