Permutations and Combinations — Page 6 | JEE Mathematics

Q51

The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is :

A 26664

B 122664

C 122234

D 22264

Correct Answer

Option A

Solution

Total possible numbers using 1, 2, 2 and 3 is =

{{4!} \over {2!}}

= 12 When unit place is 1, the total possible numbers using remaining 2, 2 and 3 are =

{{3!} \over {2!}}

= 3 When unit place is 2, the total possible numbers using remaining 1, 2 and 3 are = 3!

= 6 When unit place is 3, the total possible numbers using remaining 1, 2 and 2 are =

{{3!} \over {2!}}

= 3 $\therefore$ Sum of unit places of all (3 + 6 + 3) 12 numbers is = ( 1 $\times$ 3 + 2 $\times$ 6 + 3 $\times$ 3) Similarly, When 10th place is 1, the total possible numbers using remaining 2, 2 and 3 are =

{{3!} \over {2!}}

= 3 When 10th place is 2, the total possible numbers using remaining 1, 2 and 3 are = 3!

= 6 When 10th place is 3, the total possible numbers using remaining 1, 2 and 2 are =

{{3!} \over {2!}}

= 3 $\therefore$ Sum of 10th places of all (3 + 6 + 3) 12 numbers is = ( 1 $\times$ 3 + 2 $\times$ 6 + 3 $\times$ 3) $\times$ 10 Similarly, Sum of 100th places of all (3 + 6 + 3) 12 numbers is = ( 1 $\times$ 3 + 2 $\times$ 6 + 3 $\times$ 3) $\times$ 100 and Sum of 1000th places of all (3 + 6 + 3) 12 numbers is = ( 1 $\times$ 3 + 2 $\times$ 6 + 3 $\times$ 3) $\times$ 1000 $\therefore$ Total sum = ( 1 $\times$ 3 + 2 $\times$ 6 + 3 $\times$ 3) + ( 1 $\times$ 3 + 2 $\times$ 6 + 3 $\times$ 3) $\times$ 10 + ( 1 $\times$ 3 + 2 $\times$ 6 + 3 $\times$ 3) $\times$ 100 + ( 1 $\times$ 3 + 2 $\times$ 6 + 3 $\times$ 3) $\times$ 1000 = (3 + 12 + 9) (1 + 10 + 100 + 1000) = 1111 $\times$ 24 = 26664

Q52

If

{}^n{P_r} = {}^n{P_{r + 1}}

and

{}^n{C_r} = {}^n{C_{r - 1}}

, then the value of r is equal to :

A 1

B 4

C 2

D 3

Correct Answer

Option C

Solution

{}^n{P_r} = {}^n{P_{r + 1}} \Rightarrow {{n!} \over {(n - r)!}} = {{n!} \over {(n - r - 1)!}}

\Rightarrow (n - r) = 1

.....(1)

{}^n{C_r} = {}^n{C_{r - 1}}

\Rightarrow {{n!} \over {r!(n - r)!}} = {{n!} \over {(r - 1)!(n - r + 1)!}}

\Rightarrow {1 \over {r(n - r)!}} = {1 \over {(n - r + 1)(n - r)!}}

\Rightarrow n - r + 1 = r

\Rightarrow n + 1 = 2r

..... (2) From (1) and (2),

2r - 1 - r = 1 \Rightarrow r = 2

Q53

Let P1, P2, ......, P15 be 15 points on a circle. The number of distinct triangles formed by points Pi, Pj, Pk such that i +j + k

\ne

15, is :

A 12

B 419

C 443

D 455

Correct Answer

Option C

Solution

Total number of triangles =

{}^{15}{C_3}

i + j + k = 15 (Given) Number of possible triangles using the vertices Pi, Pj, Pk such that i + j + k $\ne$ 15 is equal to

{}^{15}{C_3}

$-$ 12 = 443 Option (c)

Q54

The number of ways to distribute 30 identical candies among four children C1, C2, C3 and C4 so that C2 receives at least 4 and at most 7 candies, C3 receives at least 2 and at most 6 candies, is equal to :

A 205

B 615

C 510

D 430

Correct Answer

Option D

Solution

By multinomial theorem, no. of ways to distribute 30 identical candies among four children C1, C2 and C3, C4 = Coefficient of x30 in (x4 + x5 + .... + x7) (x2 + x3 + .... + x6) (1 + x + x2 ....)

2 = Coefficient of x24 in

{{(1 - {x^4})} \over {1 - x}}{{(1 - {x^5})} \over {1 - x}}{{{{(1 - {x^{31}})}^2}} \over {{{(1 - x)}^2}}}

= Coefficient of x24 in

(1 - {x^4} - {x^5} + {x^9}){(1 - x)^{ - 4}}

= {}^{27}{C_{24}} - {}^{23}{C_{20}} - {}^{22}{C_{19}} + {}^{18}{C_{15}} = 430

Q55

The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is :

A 36

B 48

C 60

D 72

Correct Answer

Option D

Solution

To make a no. divisible by 3 we can use the digits 1,2,5,6,7 or 1,2,3,5,7.

Using 1,2,5,6,7, number of even numbers is = 4 × 3 × 2 × 1 × 2 = 48 Using 1,2,3,5,7, number of even numbers is = 4 × 3 × 2 × 1 × 1 = 24 Required answer is 72.

Q56

The value of

\dfrac{1}{1 ! 50 !}+\dfrac{1}{3 ! 48 !}+\dfrac{1}{5 ! 46 !}+\ldots .+\dfrac{1}{49 ! 2 !}+\dfrac{1}{51 ! 1 !}

is :

A

\dfrac{2^{51}}{50 !}

B

\dfrac{2^{51}}{51 !}

C

\dfrac{2^{50}}{50 !}

D

\dfrac{2^{50}}{51 !}

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{S}=\frac{1}{1 ! 50 !}+\frac{1}{3 ! 48 !}+\frac{1}{5 ! 46 !}+\ldots \ldots+\frac{1}{49 ! 2 !}+\frac{1}{51 ! 1 !} \\\\ & =\frac{1}{51 !}\left(\frac{51 !}{1 ! 50 !}+\frac{51 !}{3 ! 48 !}+\frac{51 !}{5 ! 46 !}+\ldots . .+\frac{51 !}{49 ! 2 !}+\frac{51 !}{51 ! 0 !}\right) \\\\ & =\frac{1}{51 !}\left({ }^{51} C_{50}+{ }^{51} C_{48}+{ }^{51} C_{46}+\ldots \ldots .+{ }^{51} C_2+{ }^{51} C_0\right) \\\\ & \because{ }^n C_0+{ }^n C_2+{ }^n C_4+\ldots \ldots=2^{n-1} \\\\ & \therefore S=\frac{2^{50}}{51 !} \end{aligned}

Q57

The letters of the word OUGHT are written in all possible ways and these words are arranged as in a dictionary, in a series. Then the serial number of the word TOUGH is :

A 79

B 84

C 89

D 86

Correct Answer

Option C

Solution

Lets arrange the letters of OUGHT in alphabetical order. G, H, O, T, U Words starting with

\begin{aligned} & \mathrm{G}----\rightarrow 4 ! \\\\ & \mathrm{H}----\rightarrow 4 ! \\\\ & \mathrm{O}----\rightarrow 4 ! \end{aligned}

$\mathrm{T} \,\mathrm{G}---\rightarrow 3$ !

T H $---\rightarrow 3$ !

T O $\mathrm{G}--\rightarrow 2$ !

T O H $--\rightarrow 2$ !

T O U G H $\rightarrow 1$ !

____________________________ Total = 89

Q58

The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48, is :

A 400

B 472

C 507

D 432

Correct Answer

Option D

Solution

Number divisible by 3 = 300 Number divisible by 4 = 225 Number divisible by 12 = 75 Number divisible by 48 = 18 Total required number = 300 + 225 $-$ 75 $-$ 18 = 432 $\therefore$ Option (1) is correct.

Q59

The number of integers, greater than 7000 that can be formed, using the digits 3, 5, 6, 7, 8 without repetition is :

A 48

B 120

C 168

D 220

Correct Answer

Option C

Solution

Four digit numbers greater than 7000 $=2 \times 4 \times 3 \times 2=48$ Five digit number $=5 !=120$ Total number greater than 7000 $=120+48=168$

Q60

The total number of three-digit numbers, divisible by 3, which can be formed using the digits

1,3,5,8

, if repetition of digits is allowed, is :

A 21

B 22

C 18

D 20

Correct Answer

Option B

Solution

The number of three-digit numbers divisible by 3 by considering the possible sums of digits that are divisible by 3.

Your approach is as follows: 1.

Sum of digits is 3: $(1, 1, 1)$ - 1 possible number 2.

Sum of digits is 9: $(1, 3, 5)$ and $(3, 3, 3)$ - Let's consider the cases separately : a.

Sum of digits is 9: $(1, 3, 5)$ For this case, we can arrange the digits in $3!$ ways : 135, 153, 315, 351, 513, and 531. b.

Sum of digits is 9: $(3, 3, 3)$ For this case, since all the digits are the same, there is only 1 possible number : 333.

Now, the total number of possible numbers when the sum of digits is 9 is :

3! + 1 = 6 + 1 = 7

3.

Sum of digits is 12: $(1, 3, 8)$ - $3!$ possible numbers 4.

Sum of digits is 15: $(5, 5, 5)$ - 1 possible number 5.

Sum of digits is 18: $(5, 5, 8)$ - $\dfrac{3!}{2!}$ possible numbers (since 5 is repeated) 6.

Sum of digits is 21: $(5, 8, 8)$ - $\dfrac{3!}{2!}$ possible numbers (since 8 is repeated) 7.

Sum of digits is 24: $(8, 8, 8)$ - 1 possible number Adding up the possible numbers for each case, we get:

1 + 7 + 3! + 1 + \frac{3!}{2!} + \frac{3!}{2!} + 1 = 1 + 7 + 6 + 1 + 3 + 3 + 1 = 22

So, there are a total of 22 three-digit numbers divisible by 3 that can be formed using the digits $1, 3, 5, 8$ with repetition allowed.