Test: JEE Main 35 Year PYQs- Permutations & Combinations - JEE MCQ

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

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

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

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

ⁿ C_{r +1} + ⁿ C_{r -1}+ 2ⁿC_r = ⁿC_{r -1}+ ⁿ C_r + ⁿC_r + ⁿC_r +1  

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.

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!

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

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 -1}C_{r -1}
∴ The required number of ways

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

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

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

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.

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

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

Total number of ways = ³C₂ x ⁹C₂

 = 3 x 36 = 108

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-1C_{r -1} .
Hence required number of ways = ^10-1C_4-1 = ⁹C₃
∴  Both statements are correct and second statement is a correct explanation of statement -1.

Number of required triangles

=  ¹⁰C₃ – ⁶C₃

= 120 - 20 =100

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 ]

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

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.

ALLMS
No. of words starting with