Permutations and Combinations — Page 11 | JEE Mathematics

Q101

The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is :

A 24

B 30

C 36

D 48

Correct Answer

Option B

Solution

Here number should be divisible by 3, that means sum of numbers should be divisible by 3.

Possible 4 digits among 0, 1, 2, 3, 4 which are divisible by 3 are (1)

\,\,\,\,

(0, 2, 3, 4) Sum of digits = 0 + 2 + 3 +4 = 9 (divisible by 3) (2)

\,\,\,\,

(0, 1, 2, 3) Sum of digits = 0 + 1 + 2 + 3 = 6 (divisible by 3) Case 1 : When 4 digits are (0, 2, 3, 4) then

\therefore\,\,\,\,

Total possible numbers =

^3{C_1}

$\times$

^3{C_1}

$\times$

^2{C_1}

$\times$

^1{C_1}

= 3 $\times$ 3 $\times$ 2 $\times$ 1 = 18 Case 2 : When 4 digits are (0, 1, 2, 3) then,

\therefore\,\,\,\,

Total possible number in this case =

^2{C_1}

$\times$

^3{C_1}

$\times$

^2{C_1}

$\times$

^1{C_1}

= 2 $\times$ 3 $\times$ 2 $\times$ 1 = 12

\therefore\,\,\,\,

Total possible numbers will be = 18 + 12 = 30

Q102

If

\sum\limits_{r = 0}^{25} {\left\{ {{}^{50}{C_r}.{}^{50 - r}{C_{25 - r}}} \right\} = K\left( {^{50}{C_{25}}} \right)} ,\,\,

then K is equal to :

A 224

B 225

-

1

C 225

D (25)2

Correct Answer

Option C

Solution

\sum\limits_{r = 0}^{25} {^{50}} {C_r}.{}^{50 - r}{C_{25 - r}}

= \sum\limits_{r = 0}^{25} {{{50!} \over {r!\left( {50 - r} \right)!}}} \times {{\left( {50 - r} \right)!} \over {\left( {25} \right)!\left( {25 - r} \right)!}}

= \sum\limits_{r = 0}^{25} {{{50!} \over {25!25!}} \times {{25!} \over {\left( {25 - r} \right)!\left( {r!} \right)}}}

= {}^{50}{C_{25}}\sum\limits_{r = 0}^{25} {^{25}} {C_r} = \left( {{2^{25}}} \right){}^{50}{C_{25}}

$\therefore$

K = {2^{25}}

Q103

If

{}^n{C_r}

denotes the number of combination of n things taken r at a time, then the expression

\,{}^n{C_{r + 1}} + {}^n{C_{r - 1}} + 2\, \times \,{}^n{C_r}

equals

A

\,{}^{n + 1}{C_{r + 1}}

B

{}^{n + 2}{C_r}

C

{}^{n + 2}{C_{r + 1}}

D

\,{}^{n + 1}{C_r}

Correct Answer

Option C

Solution

Arrange it this way,

^n{C_{r + 1}} + 2.{}^n{C_r} + {}^n{C_{r - 1}}

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

\left[ \, \right.

Now use the rule,

\left. {{}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}} \right]

= {}^{n + 1}{C_{r + 1}} + {}^{n + 1}Cr

= {}^{n + 2}{C_{r + 1}}

Q104

\sum\limits_{k = 0}^6 {{}^{51 - k}{C_3}}

is equal to :

A

\mathrm{{}^{51}{C_4} - {}^{45}{C_4}}

B

\mathrm{{}^{51}{C_3} - {}^{45}{C_3}}

C

\mathrm{{}^{52}{C_3} - {}^{45}{C_3}}

D

\mathrm{{}^{52}{C_4} - {}^{45}{C_4}}

Correct Answer

Option D

Solution

\begin{aligned} & \sum_{\mathrm{k}=0}^6{ }^{51-\mathrm{k}} \mathrm{C}_3 \\\\ & ={ }^{51} \mathrm{C}_3+{ }^{50} \mathrm{C}_3+{ }^{49} \mathrm{C}_3+\ldots+{ }^{45} \mathrm{C}_3 \\\\ & ={ }^{45} \mathrm{C}_3+{ }^{46} \mathrm{C}_3+\ldots \ldots+{ }^{51} \mathrm{C}_3 \\\\ & ={ }^{45} \mathrm{C}_4+{ }^{45} \mathrm{C}_3+{ }^{46} \mathrm{C}_3+\ldots . .+{ }^{51} \mathrm{C}_3-{ }^{45} \mathrm{C}_4 \\\\ & \left({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}={ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}}\right) \\\\ & ={ }^{52} \mathrm{C}_4-{ }^{45} \mathrm{C}_4 \end{aligned}

Q105

The value of (2.1P0 – 3.2P1 + 4.3P2 .... up to 51th term) + (1! – 2! + 3! – ..... up to 51th term) is equal to :

A 1

B 1 + (51)!

C 1 – 51(51)!

D 1 + (52)!

Correct Answer

Option D

Solution

(2.1P0 – 3.2P1 + 4.3P2 .... up to 51th term) + (1! – 2!

+ 3! – ..... up to 51th term) = ( 2.1! - 3.2!

+ 4.3! - ....

+ 52.51!)

+ ( 1! – 2!

+ 3! – .....

+ (51)!

) = (2! - 3!

+ 4! .....

+ 52!

) + ( 1! - 2!

+ 3! - 4!

+ .....

+ 51!

) = 1 + (52)!

Q106

If the letters of the word MATHS are permuted and all possible words so formed are arranged as in a dictionary with serial numbers, then the serial number of the word THAMS is :

A 103

B 104

C 102

D 101

Correct Answer

Option A

Solution

To solve this problem, we start by finding the number of permutations before we reach a word that begins with T.

We have 5 letters in the word MATHS.

If we fix the first letter, there are 4!

(=4 $\times$ 3 $\times$ 2 $\times$ 1=24) ways to arrange the remaining letters.

The letters before T in alphabetical order are A, H, M, and S.

For each of these 4 letters, there are 4!

ways to arrange the rest of the word, giving 4 $\times$ 4!

= 96 words that come before any words starting with T.

Then, in the section of words that start with 'T', we start with 'TA'.

There are 3 remaining letters after 'TA', hence the permutations for words that start with 'TA' are (5-2)!

=3!

=6.

So, the total permutations for all words that come before 'TH' is 96 + 6 = 102.

Now, we need to count the words that come before 'THAMS'.

After 'TH', the letters left are 'AMS'.

Arranged in dictionary order, they would be 'AMS', 'ASM', 'MAS', 'MSA', 'SAM', 'SMA'.

So, 'THAMS' is the first word in the 'TH' category, so it is the 103rd word overall when counting from the beginning.

So, the correct answer is 103, which corresponds to Option A.

Q107

The number of square matrices of order 5 with entries from the set {0, 1}, such that the sum of all the elements in each row is 1 and the sum of all the elements in each column is also 1, is :

A 125

B 150

C 225

D 120

Correct Answer

Option D

Solution

$\because$ In every row and every column there would be exactly one 1 and four zeroes.

Number of matrices $={ }^{5} C_{1} \cdot{ }^{4} C_{1} \cdot{ }^{3} C_{1} \cdot{ }^{2} C_{1} \cdot{ }^{1} C_{1}=$ 120

Q108

There are 5 points

P_1, P_2, P_3, P_4, P_5

on the side

A B

, excluding

A

and

B

, of a triangle

A B C

. Similarly there are 6 points

\mathrm{P}_6, \mathrm{P}_7, \ldots, \mathrm{P}_{11}

on the side

\mathrm{BC}

and 7 points

\mathrm{P}_{12}, \mathrm{P}_{13}, \ldots, \mathrm{P}_{18}

on the side

\mathrm{CA}

of the triangle. The number of triangles, that can be formed using the points

\mathrm{P}_1, \mathrm{P}_2, \ldots, \mathrm{P}_{18}

as vertices, is:

A 751

B 776

C 796

D 771

Correct Answer

Option A

Solution

Number of points on side

A B=5

Number of points on side

B C=6

Number of points on side

A C=7

Number of ways selecting three points from side

A B={ }^5 C_3

Number of ways selecting three points from side

B C={ }^6 C_3

Number of ways selecting three points from side

A C={ }^7 C_3

Total number of triangle possible formed using the points

P_1 P_2 \ldots P_{18}

\begin{aligned} & ={ }^{18} C_3-{ }^5 C_3-{ }^6 C_3-{ }^7 C_3 \\\\ & =816-10-20-35 \\\\ & =751 \end{aligned}

Q109

The number of five digit numbers, greater than 40000 and divisible by 5 , which can be formed using the digits

0,1,3,5,7

and 9 without repetition, is equal to :

A 132

B 72

C 120

D 96

Correct Answer

Option C

Solution

Since the five-digit number must be greater than 40000, the only options for the first digit are 5, 7, or 9.

That leaves 3 remaining choices for the first digit.

Since the number has to be divisible by 5, the last digit must be 0 or 5.

If the first digit is 5, the last digit can only be 0, since digits cannot be repeated.

If the first digit is 7 or 9, the last digit can be 0 or 5, so there are 2 choices.

Five possible configurations for the first and last digits : - 5xxx0 - 7xxx0 - 7xxx5 - 9xxx0 - 9xxx5 For each of these configurations, there are three places in the middle that can be filled with the remaining 4 unused digits.

Since repetition isn't allowed, there are 4 options for the second digit, 3 for the third, and 2 for the fourth, which can be represented as 4P3, or permutations of 4 items taken 3 at a time.

This is equal to 4 $\times$ 3 $\times$ 2 = 24.

So, for each of the 5 configurations, there are 24 ways to arrange the middle three digits.

Therefore, the total number of five-digit numbers that meet the criteria is 5 $\times$ 24 = 120.