JEE Advanced Level Test: Permutation And Combinations- 2 - JEE MCQ

Required total number of ways = 

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

The required number of ways 

Total number of four digit odd numbers = 6 × 7 × 7 × 4 = 1176.

Required number of ways = ⁶C₂ × ⁵C₄ + ⁶C₃ × ⁵C₃ + ⁶C₄ × ⁵C₂ = 75 + 200 + 150 = 425

Total number of solution = 3 × 3 × 3 = 27

80 cards 5 equal groups of 16 each
⁸⁰C₁₆ * ⁶⁴C₁₆ * ⁴⁸C₁₆ * ³²C₁₆ * ¹⁶C₁₆
= 80!/(16!)⁵
But all groups are equal 
therefore, 80!/[(16!)⁵ * 5!]

Any divisor of 2⁵. 3⁴. 5² is of the form 2^a, 3^b. 5^c where  0 < a < 5, 0 < b < 4 and 0 < c < 2. 
Hence the sum of the divisors 

 (1 + 2 + . . . 2⁵) (1 + 3 + . . . + 3⁴) (1 + 5 + 5²) 

The committee of 5 has to be formed in which at least two boys and two girls is must and we have total 6 boys and 5 girls. 
So our choices are to select two boys and three girls or three boys and two girls out of six boys and five girls
Hence, required number of ways = ⁶C₂​×⁵C_2​+⁶C₄×⁵C₁
​=275

Each object can be put either in box B₁ (say) or in box B₂ (say). So, there are two choices for each of the n objects. Therefore the number of choices for n distinct objects is

One of these choices correspond to either the first or the second box being empty.
Thus, there are 2ⁿ – 2 ways in which neither box is empty.

Required number of ways = ⁸C₂ × ⁶C₂ × 2² = 1680.

2I + 3E + 1A = 6 
Set-I = 2I + 3E + 1A = 6
Set-II = NTRMPT = 6 
Set-III six letters arranged 6!/2! Ways of these 7 gaps 6 letters of set I are arranged in 

C =1
H =1
S =1
E =3
No. of ways of arranging the letters of CHEESE = 6!/3!
=720/6 = 120.
No. of ways of rearranging the letters of CHEESE =120−1
=119

The correct option is D.

Number of ways in which 8 people can sit around a round table 

= (8 — 1)! = 5040 ways 

If three girls sit together, we consider it as 1 group and these 

three girls can sit in 3! ways. 

So, number of ways in which 3 girls sit together in a 

the circular table of 8 people is 3!5! = 720. 

Required number of ways in which no three girls sit together = 

5040 -720 = 4320  

Any selection of three digits from the ten digits 0, 1, 2, 3, … 9 gives one number. It is of ¹⁰C₃ ways. 

Wearing of coats = ⁴P₃
Wearing of waist coats = ⁵P₃
Wearing of caps = ⁶P₃

2 vowels occupy 3 even places to 3P2 ways remaining 4 places occupy by consonants in 4!

⁶P₃ + ⁶P₃ – ⁵P₂ = 220 

The formula to find the number of diagonals = n(n-3)/2, where n = no. of sides of the polygon.
Here n(n-3)/2 = 35, or
n(n-3) = 70
n² - 3n - 70 = 0, or
(n-10)(n+7) = 0, lor
n = 10 or -7.
Hence it has 10 sides.

⁹C₂. ⁹C₂

MADHUR : ADHMRU 
3. 5! + 0.4! + 0.3! + 0.2! + 1.1! + 1 = 362 

Since we are given that 20 persons are invited. Therefore including Host there are total 21 people, which are supposed to be seated around a circular table.
Now there is a constraint that two particular persons are seated on either side of the host.
Therefore considering host and these two particular persons as one entity, we are left with a total of (21 - 3 + 1) persons i.e. 19 persons, which are to be seated around the circular table.
This can be done in (19 - 1)! ways
i.e. 18!
But those two particular persons who are seated on either side of host can interchange their positions i.e. they can be arranged in 2! ways.
Hence total number of ways = 18!*2!
i.e. 2*18!
Therefore with the given constraint all the invited persons including host can be seated around a circular table in total of 2*18! ways.

Total arrangement with the letters of the word DEVIL = 5! = 120
Number of arrangement starting with D = 24 = 4!
Number of arrangement end with L = 24 = 4!
Number of arrangement that begin with D and end with L is 6
Number of arrangements required = 120 – (24 + 24 – 6) = 78 

x + y + z + w = 3
Number of non-negative integral solutions ^3+4-1C_4-1 = 20

NNN = One unit
Remaining = 8 units, total 9 units
∴ Total permutations – E’s come together 

No two vowels come together
No of permutations = (7!/2!2!) * 8C4 * 4!/2!
= (7!*4!)/(2!)³ * 8!/4!4!
7!/(2!)³ * 8P4 

This is a type of question in which we have to arrange n things out of which p,q and r are similar or identical things
So in this case n is 6 and p,q,r is 2,2,2 respectively
Number of 2 is 2
Number of 3 is 2
Number of 9 is 2
∴ Required number of ways is 6!/(2!×2!×2!)
=90

SWORD can be arranged such that no letter is in its original position, that is all letter is wrongly positioned (De arrangement)
De-arrangement formula is n!(1 − 1/1 + 1/2! − 1/3! +.....(−1)ⁿ 1/n!)
In this case n = 5
Number of ways is 5!/2! − 5!/3! + 5!/4! − 1
= 44

If y = 0, then  = 1, 2, 3, 4, ….9
y = 1, then x = 2, 3, 4, ....9 and ... so on 
So on continuing the number of required number 

2160 = 2⁴. 3³. 5¹
Proper factors = (4 + 1)(3 + 1)(1 + 1) - 2 = 38 

Given number of vowels = 4 and number of consonants = 5 We have to form words by 2 vowels and 3 consonants.
So, lets first select 2 vowels and 3 consonants.
Number of ways of selection = ⁴C₂ x ⁵C₃ = 6 x 10 = 60
Now, these letters can be arranged in 5! ways.
So, total number of words = 60 x 5! = 60 x 120 = 7200