MATHS M.C.Q (1 Marks)

4. $\text{r! }^{ \text{n}-3}\text{C}_{\text{r}-3}$

Solutions: 

Here, we have to permute n things of which 3 things are to be included. So, only the remaining (n − 3) things are left for permutation, taking (r − 3) things at a time. This is because 3 things have already been included. But, these r things can be arranged in r! ways.

$\therefore$ Total number of permutations $=\text{r! }^{ \text{n}-3}\text{C}_{\text{r}-3}$ 

4. 144

Solutions: 

The word ARTICLE consists of 3 vowels that have to be arranged in the three even places. This can be done in 3! ways. And, the remaining 4 consonants can be arranged among themselves in 4! ways.

$\therefore$ Total number of ways = 3! × 4! = 144

1. 120

Solutions: 

Total number of arrangements of the letters of the word CHEESE = Number of arrangements of 6 things taken all at a time, of which 3 are of one kind $=\frac{6!}{3!}=120$ 

1. 60 × 5!

Solutions: 

The four people, i.e A, B and the two persons between them are always together. Thus, they can be considered as a single person. So, along with the remaining 4 persons, there are now total 5 people who need to be arranged. This can be done in 5! ways. But, the two persons that have to be included between A and B could be selected out of the remaining 6 people in ⁶P₂ ways, which is equal to 30. For each selection, these two persons standing between A and B can be arranged among themselves in 2 ways.

$\therefore$ Total number of arrangements = 5! × 30 × 2 = 60 × 5!

3. 2¹² − 1

Solutions: 

Each of the bulb has its own switch, i.e each bulb will have two outcomes − it will either glow or not glow. Thus, each of the 12 bulbs will have 2 outcomes.

$\therefore $ Total number of ways to illuminate the room = 2¹²

Here, we have also considered the way in which all the bulbs are switched-off. However, this is not required as we need to find out only the number of ways of illuminating the room.

Hence, we subtract that one way from the total number of ways.

= 2¹² − 1

3. 36

Solutions: 

The word APURBA is a 6 letter word consisting of 3 vowels that can be arranged in 3 alternate places, in $\frac{3!}{2!}$ ways.

The remaining 3 consonants can be arranged in the remaining 3 places in 3! ways.

$\therefore$ Total number of words that can be formed $=\frac{3!}{2!}\times3!=18$

But this whole arrangement can be set-up in total two ways, i.e either VCVCVC or CVCVCV.

$\therefore$ T otal number of words = 18 × 2 = 36

1. 72

Solutions: 

When we make words after selecting letters of the word BHARAT, it could consist of a single A, two As or no A.

Case-I: A is not selected for the three letter word.

Number of arrangements of three letters out of B, H, R and T = 4 × 3 × 24 × 3 × 2 = 24

Case-II: One A is selected and the other two letters are selected out of B, H, R or T. Possible ways of selection: Selecting two letters out of B, H, R or T can be done in ⁴P₂ = 12 ways. Now, in each of these 12 ways, these two letters can be placed at any of the three places in the three letter word in 3 ways.

$\therefore$ Total number of words that can be formed = 12 × 3 = 36

Case-III: Two A's and a letter from B, H, R or T are selected.

Possible ways of arrangement: 

Number of ways of selecting a letter from B, H, R or T = 4 And now this letter can be placed in any one of the three places in the three letter word other than the two A's in 3 ways.

$\therefore$ Total number of words having 2 A's = 4 × 3 = 12

Hence, total number of words that can be formed = 24 + 36 + 12 = 72

2. 360

Solutions: 

10 lakhs consists of seven digits.

 Number of arrangements of seven numbers of which 2 are similar of first kind, 3 are similar of second kind $=\frac{7!}{2!\ 3!}$ 

But, these numbers also include the numbers in which the first digit has been considered as 0. This will result in a number less than 10 lakhs. Thus, we need to subtract all those numbers.

Numbers in which the first digit is fixed as 0 = Number of arrangements of the remaining 6 digits $=\frac{6!}{2!\ 3!}$ 

Total numbers greater than 10 lakhs that can be formed using the given digits $=\frac{7!}{2!\ 3!}-\frac{6!}{2!\ 3!}$

$=420-60$

$=360$

1. 24

Solutions: 

In order to make a number divisible by 4, its last two digits must be divisible by 4, which in this case can be 12, 24, 32 or 52.

Since repetition of digits is not allowed, the remaining first two digits can be arranged in 3 × 2 ways in each case.

$\therefore$ Total number of numbers that can be formed = 4 × {3 × 2} = 24

1. 216

Solutions: 

A number is divisible by 3 when the sum of the digits of the number is divisible by 3. Out of the given 6 digits, there are only two groups consisting of 5 digits whose sum is divisible by 3.

= 1 + 2 + 3 + 4 + 5 = 15

= 0 + 1 + 2 + 4 + 5 = 12

Using the digits 1, 2, 3, 4 and 5, the 5 digit numbers that can be formed = 5! Similarly, using the digits 0, 1, 2, 4 and 5, the number that can be formed = 5! - 4! {since the first digit cannot be 0}

$\therefore$ Total numbers that are possible = 5! + 5! - 4! = 240 - 24 = 216

1. 12

Solutions: 

All S's can be placed either at even places or at odd places, i.e. in 2 ways. The remaining letters can be placed at the remaining places in 3!, i.e. in 6 ways.

$\therefore$ Total number of ways = 6 × 2 × 2 = 12

2. 240

Solutions: 

Total number of words that can be formed of the letters of the word BHARAT $=\frac{6!}{2!}$ 

Number of words in which the letters B and H are always together $=2\times\frac{5!}{2!}$

$=120$

$\therefore$ Number of words in which the letters B and H are never together $=360-120$

$=240$

1. 360

Solutions: 

The word CONSTANT consists of two vowels that are placed at the 2nd and 6th position, and six consonants.

The two vowels can be arranged at their respective places, i.e. 2nd and 6th place, in 2! ways.

The remaining 6 consonants can be arranged at their respective places in $\frac{6!}{2!\ 2!}$ ways.

$\therefore$ Total number of arrangements $=2!\times\frac{6!}{2!\ 2!}=360$

1. 6 and 17

Solutions: 

$^\text{k}+5\text{P}_\text{k+1}=\frac{11(\text{k}-3)}{2}.\ ^\text{k+3}\text{P}_\text{k}$

$\Rightarrow \frac{(\text{k}+5)!}{(\text{k}+5-\text{k}-1)!​​}=\frac{11(\text{k}-1)}{2}\times \frac{(\text{k}+3)!}{(\text{k}+3-\text{k})!}$

$\Rightarrow \frac{(\text{k}+5)!}{(\text{k}+3)!}=\frac{11(\text{k}-1)}{2}\times\frac{4!}{3!}$

$\Rightarrow (\text{k}+5)(\text{k}+4)=22(\text{k}-1)$

$\Rightarrow \text{k}^2+9\text{k}+20=22\text{k}-22$

$\Rightarrow \text{k}^2-13\text{k}+42=0$

$\Rightarrow \text{k}=6,7$

4. 69760

Solutions: 

Total number of five digit numbers (since there is no restriction of the number 0XXXX) = 10 × 10 × 10 × 10 × 10 = 100000.

These numbers also include the numbers where the digits are not being repeated. So, we need to subtract all such numbers.

Number of 5 digit numbers that can be formed without any repetition of digits = 10 × 9 × 8 × 7 × 6 = 30240

$\therefore$ Number of five-digit telephone numbers having at least one of their digits repeated = {Total number of 5 digit numbers} - {Number of numbers that do not have any digit repeated} = 100000 - 30240 = 69760

1. 576

Solutions: 

There are 3 even places in the 7 letter word ARTICLE. So, we have to arrange 4 consonants in these 3 places in ⁴P₃ ways. And the remaining 4 letters can be arranged among themselves in 4! ways.

$\therefore$ Total number of ways of arrangement = ⁴P₃ × 4! = 4! × 4! = 576

2. 1956

Solutions: 

Number of permutations of six signals taking 1 at a time = ⁶P₁ 

Number of permutations of six signals taking 2 at a time = ⁶P₂

Number of permutations of six signals taking 3 at a time = ⁶P₃

Number of permutations of six signals taking 4 at a time = ⁶P₄

Number of permutations of six signals taking 5 at a time = ⁶P₅

Number of permutations of six signals taking all at a time = ⁶P₆

_$\therefore$ Total number of signals

$=\frac{6!}{5!}+\frac{6!}{4!}+\frac{6!}{3!}+\frac{6!}{2!}+\frac{6!}{1!}+6!$

$=6+30+120+360+720+720$

$=1956$

1. 4! × 3!

Solutions: 

According to the question, 3 men have to be 'consecutive' means that they have to be considered as a single man. But, these 3 men can be arranged among themselves in 3! ways. And, the remaining 3 men, along with this group, can be arranged among themselves in 4! ways.

$\therefore$ Total number of arrangements = 4! × 3!

1. 324

Solutions: 

When arranged alphabetically, the letters of the word KRISNA are A, I, K, N, R and S.

Number of words that will be formed with A as the first letter = Number of arrangements of the remaining 5 letters = 5!

Number of words that will be formed with I as the first letter = Number of arrangements of the remaining 5 letters = 5!

$\therefore$ The number of words beginning with KA = Number of arrangements of the remaining 4 letters = 4!

The number of words beginning with KA = Number of arrangements of the remaining 4 letters = 4!

The number of words starting with KN = Number of arrangements of the remaining 4 letters = 4!

Alphabetically, the next letter will be KR.

Number of words starting with KR followed by A, i.e. KRA = Number of arrangements of the remaining 3 letters = 3!

Number of words starting with KRI followed by A, i.e. KRIA = Number of arrangements of the remaining 2 letters = 2!

Number of words starting with KRI followed by N, i.e. KRIN = Number of arrangements of the remaining 2 letters = 2!

The first word beginning with KRIS is the word KRISAN and the next word is KRISNA.

$\therefore$ Rank of the word KRISNA = 5! + 5! + 4! + 4! + 4! + 3! + 2! + 2! + 2

= 324

3. 6

Solutions: 

According to the question:

$^\text{n}\text{P}_4=12\times\ ^\text{n}\text{P}_2$

$\Rightarrow \frac{\text{n!}}{(\text{n}-4)!}=12\times\frac{\text{n!}}{\text{(n-2)!}}$

$\Rightarrow \frac{(\text{n}-2)!}{(\text{n}-4)!}=12$

$\Rightarrow (\text{n}-2)(\text{n}-3)=4\times 3$

$\Rightarrow \text{n}-2=4$

$\Rightarrow \text{n}=6$