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Maths 2 Marks Questions

Permutations · CBSE STD 11 Science (CBSE - English Medium). 68 questions.

Questions · Page 2 of 2

2 Marks Questions

Question 512 Marks
How many words can be formed from the letters of the word 'SUNDAY'? How many of these begin with D?
Answer
There are $6$ letters in the word 'SUNDAY'. The total number of words formed with these 6 letters is the number of arrangements of $6$ items, taken all at a time, which is equal to $^6P_6 = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720.$
If we fix up Din the beginning, then the remaining $5$ letters can be arranged in $^5P_5 = 5!$ ways.
so, the total number of words which begin with $D = 5!$
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Question 522 Marks
How many 6-digit telephone numbers can be constructed with digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each number starts with 35 and no digit appears more than once?
Answer
Total number of digits = 10 The first two digits of telephone is 35 and no digit appears more than once.$\therefore$ Total number of remaining digits = 10 - 2 = 8
And, Total number of remaining digits of telephone number = 6 - 2 = 4 $\therefore$ Required number of telephone numbers $=\ ^8\text{P}_4$ $=\frac{8!}{(8-4)!}$ $=\frac{8!}{4!}$ $=\frac{8\times7\times6\times5\times4!}{4!}$ $= 1680$
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Question 532 Marks
How many words can be formed out of the letters of the word 'ARTICLE', so that vowels occupy even places?
Answer
We have to arrange 7 letters in a row such that vowels occupy even places. There are 3 even places (2, 4, 6). Three vowels can be arranged in these 3 even places in 3! ways. Remaining 4 odd places (1, 3, 5, 7) are to be occupied by the 4 consonants. This can be done in 4! ways.
Hence, the total number of words in which vowels occupy even places = 3! × 4!
= 3 × 2 × 4 × 3 × 2 = 144
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Question 542 Marks
There are 6 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next three have 2 each?
Answer
Each one of the first three questions can be answered in 4 ways. $\therefore$ The total number of ways to answered the first three question = 4 × 4 × 4 = 64 Each of the next three question can be answered in 2 ways. $\therefore$ The total number of ways the answered the next three qusstions = 2 × 2 × 2 = 8so, total number of sequences at answers = 64 × 8 = 512
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Question 552 Marks
How many numbers of six digits can be formed from the digits 0, 1, 3, 5, 7 and 9 when no digit is repeated? How many of them are divisible by 10?
Answer
Total number of digits = 6
we cannot have Oat the first digit of the required six-digit numbers
The digits cannot repeat in the six digits number.
$\therefore$ Total number of six digit number are = 5 × 5 × 4 × 3 × 2 × 1 = 600
Now, the six digit number can be divided by 10, if its last digit is 0.
$\therefore$ Total numbers which are divisible by 10 = 5 × 4 × 3 × 2 × 1 × 1 = 120
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Question 562 Marks
A team consists of 6 boys and 4 girls and other has 5 boys and 3 girls. How many single matches can be arranged between the two teams when a boy plays against a boy and a girl plays against a girl?
Answer
A boy can be selected from the first team in 6 ways, and from the second in 5 ways.
so, number of single matches between the boys of two teams = 6 × 5 = 30.
similarly, the number of single matches between the girls of two teams = 4 × 3 = 12.
so, total number of matches = 30 + 12 = 42.
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Question 572 Marks
How many A.P.'s with 10 terms are there whose first term is in the set {1, 2, 3} and whose common difference is in the set {1, 2, 3, 4, 5}?
Answer
There are 3 ways to choose the first form and corresponding to each such way there are 5 ways of selecting the comm on difference.
So, required number of A.P.'s
= 3 × 5
= 15
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Question 582 Marks
In how many ways can $6$ boys and $5$ girls be arranged for a group photograph if the girls are to sit on chairs in a row and the boys are to stand in a row behind them?
Answer
Total number of boys $= 6$
Total number of girls $= 5$
Now,
Five girls can sit on chairs in a row in  $^5P_5 = 5$! ways.
and 6 boys can stand behind them in a row in  $^6P_6 = 6!$ ways.
Hence, the total number of ways
$= 5! \times 6!$
$= 5 \times 4 \times 3 \times 2 \times 1 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$= 120 \times 720$
$= 86400$
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Question 592 Marks
A person wants to buy one fountain pen, one ball pen and one pencil from a stationery shop. If there are $10$ fountain pen varieties, $12$ ball pen varieties and $5$ pencil varieties, in how many ways can he select these articles?
Answer
Here the person is to perform three jobs.
  1. Selection a ball pen from $12$ ball pens.
  2. Selection a fountain pen form $10$ fountain pens.
  3. Selection a pencil form $5$ pencils.
The first of these can be perfomed in $12$ ways, the second in $10$ ways and thrid in $5$ ways.
Therefore by the fundamental pricipal of multipilcation, the requied number of ways is $12 \times 10 \times 5 = 600$
Hence, the person can make the selection of a fountain pen, ball pen and pencil in $600$ ways.
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Question 602 Marks
Four books, one each in Chemistry, Physics, Biology and Mathematics, are to be arranged in a shelf. In how many ways can this be done?
Answer
Total number of books = 4 $\therefore$ Total number of ways = Number of arrangments of 4 books, taken all at a time $= \ ^{4}\text{P}_4$$=\frac{4!}{(4-4)!} \ \bigg[\because ^\text{n}\text{p}_\text{r}=\frac{\text{n!}}{(\text{n-r})!}\bigg]$
$=\frac{4!}{0!}$
$=4! \ [\because 0!=1]$
$=4\times3\times2\times1$
$=24$
Hence, the total number of ways to arrange the books in a shelf = 24
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Question 612 Marks
Find x in the following.
$\frac{1}{4!}+\frac{1}{5!}=\frac{\text{x}}{6!}$
Answer
We have,
$\frac{1}{4!}+\frac{1}{5!}=\frac{\text{x}}{6!}$
$\Rightarrow​​​​​​​​\frac{1}{4!}+\frac{1}{4\times5!}=\frac{\text{x}}{6\times5\times4!}$
$\Rightarrow4!\times\bigg[\frac{1}{4!}+\frac{1}{5\times4!}\bigg]=\frac{\text{x}}{30}$
$\Rightarrow1+\frac{1}{58}=\frac{\text{x}}{30}$
$\Rightarrow\frac{6}{5}=\frac{\text{x}}{30}$
$\Rightarrow\frac{\text{x}}{30}=\frac{6}{5}$
$\Rightarrow\text{x}=\frac{6\times30}{5}$
$\Rightarrow6\times6$
$\Rightarrow\text{x}= 36$
Hence, $\text{x}= 36$
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Question 622 Marks
How many different numbers of six digits can be formed from the digits 3, 1, 7, 0, 9, 5 when repetition of digits is not allowed?
Answer
We cannot have 0 at the first dig it of six-digit numbers.
So, the first digit of six-digit numbers can be selected in 5 ways.
Now, 5 digits are left including 0. So, second digit of six-digit numbers can be selected in 5 ways.
Third digit of six-diqit numbers can be selected in 4 ways.
Fourth digit of six-digit numbers can be selected in 3 ways
Fifth digit of six-digit numbers can be selected in 2 ways.
Last digit of six-digit numbers can be selected in 1 ways.
Hence, total number of num bers = 5 × 5 × 4 × 3 × 2 × 1 = 600
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Question 632 Marks
How many different five$-$digit number licence plates can be made if.
  1. First digit cannot be zero and the repetition of digits is not allowed.
  2. The first$-$digit cannot be zero, but the repetition of digits is allowed?
Answer
  1. Zero cannot be first digit of the license plates.
This means the first digit can be selected from the $9$ digits $1, 2, 3, 4 .... , 9$
So, there are $9$ ways of filling the first digit of the license plates.
Now, $9$ digits are Ieft including $0.$
So, second place can be filled with any of the remaining $9$ digits in $9$ ways.
The third place of the license plates can be filled with in any of the remaining $8$ digits.
So, there are $8$ ways of filling the third place.
The fourth place of the license plates can be filled with in any of the remaining $7$ digits.
So, there are $7$ ways at filling the fourth place.
The last place of the license plates can be filled with in any of the remaining $6$ digits.
So, there are $6$ ways of filling the fourth place.
Hence, the total number of ways $= 9 \times 9 \times 8 \times 7 \times 6 = 27216$
  1. Zero cannot be first digit of the license plates.
$\therefore$ First digit can be selected from the $9$ digits $1, 2, 3 ..... , 9$
​​​​​​​So, there are $9$ ways at filling the first digit of the licence plates.
The repetition of digits is allowed to made a license plates number.
$\therefore$ The number of ways to fill the remaining places of the number platas $= 10 \times 10 \times 10 \times 10.$
Hence, the total number of ways $= 9 \times 10 \times 10 \times 10 \times 10 = 90,000.$
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Question 642 Marks
A mint prepares metallic calendars specifying months, dates and days in the form of monthly sheets $($one plate for each month$).$ How many types of calendars should it prepare to serve for all the possibilities in future years?
Answer
The mint has to perform two job,
  1. Selecting the number of days in the february month $($there can be $28$ days or $29$ days$).$
  2. Selecting the first day of february.
The first job can be compeleted in $2$ ways the second can be performed in $7$ ways by selecting are on of the seven days of a week.
Thus, the required number of plates $= 2 \times 7 = 14$
Hence, total number of calendars $= 7 \times 2 = 14$
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Question 652 Marks
Evaluate the following:P(6,4)
Answer
We have,
${P}(6,4) = \frac{6!}{(6-4)!}\Big[\because^\text{n}\text{P}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$
$=\frac{6!}{2!}$
$= \frac{6\times5\times4\times3\times2\times1}{2!}$
$= 360$
Hence, $\text{P}(6,4)= 720$
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Question 662 Marks
How many four digit different numbers, greater than 5000 can be formed with the digits 1, 2, 5, 9, 0 when repetition of digits is not allowed?
Answer
Since the required num bars are greater than 5000.
$\therefore$ the thousand's place can be filled with any of two digits 5 or 9.
So, there are 2 ways of filling the thousand's place.
Since repetition of digits is not allowed, so the hundred's ten's and one's places can be filled in 4, 3 and 2 ways respectively.
Hence, the required number of num bers = 2 × 4 × 3 × 2 = 49
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Question 672 Marks
In how many ways can the letters of the word 'FAILURE' be arranged so that the consonants may occupy only odd positions?
Answer
There are $4$ vowels and $3$ consonants in the word 'FAILURE'.
We have to arrange $7$ letters in a row such that consonants occupy odd places. There are $4$ odd places $(1, 3, 5, 7).$ There consonants can be arranged in these $4$ odd places in $^4P_3 $ ways.
Remaining $3$ even places $(2, 4, 6)$ are to be occupied by the 4 vowels. This can be done in $^4P_3$ ways.
Hence, the total number of words in which consonants occupy odd places $= \ ^4\text{P}_3 \times\ ^4\text{P}_3$
$= \frac{4!}{(4-3)!} \times \frac{4!}{(4-3)!}$
$= 4\times3\times 2\times 1\times 4\times3\times 2\times 1$
$=24\times24$
$=576$
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Question 682 Marks
Compute:
L.C.M. (6!,7!,8!)
Answer
We have,
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
6! = 6 × 5 × 4 × 3 × 2 × 1
and
$\therefore$ L.C.M. (6!,7!,8!) = 8!
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2 Marks Questions - Page 2 - Maths STD 11 Science Questions - Vidyadip