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Maths 2 Marks Questions

Combinations · Jharkhand Board (JAC) STD 11 Science (Jharkhand - English Medium). 10 questions.

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Question 12 Marks
Evaluate the following:$\ ^{12}C_{10}$
Answer
We have,
$=\frac{12!}{10!(12-10)!}$
$=\frac{12\times11\times10!}{10!\times2\times1}$
$=66$
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Question 32 Marks
There are $10$ persons named $P_1, P_2, P_3, ...., P_{10}.$ Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement $P_1$  must occur whereas $P_4$ and $P_5$ do not occur. Find the number of such possible arrangements.
Answer
Total persons $= 10$
Number of persons to be selected $= 5$
Condition $= P_1 $ must and $P_4, P_5 $ must not be there.
Remaining number of person required is $4$ out of $10 - 3 = 7$
$={^\text{7}}\text{C}_{\text{4}}\times5!$
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Question 42 Marks
From a group of 15 cricket players, a team of 11 players is to be chosen. In how many ways can this be done?
Answer
Required number of ways $={^\text{15}\text{C}}_{11}$
Now,
$\Rightarrow {^\text{15}\text{C}}_{11}={^\text{15}\text{C}}_{4}$
$=\frac{15}{4}\times\frac{14}{3}\times\frac{13}{2}\times\frac{12}{1}\times{^\text{11}}\text{C}_{0}$
$=1365$
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Question 52 Marks
Evaluate the following:
$^{14}C_3$
Answer
We have,
$=\frac{14!}{3!(14-3)!}$
$=\frac{14!}{3!11!}$
$=\frac{14\times13\times\times12\times11!}{3\times2\times1\times11!}$
$=\frac{14\times13\times12}{6}$
$=364$
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Question 62 Marks
How many words, with or without meaning can be formed from the letters of the word $\text{'MONDAY',}$ assuming that no letter is repeated, if:
  1. $4$ letters are used at a time.
  2. All letters are used at a time.
  3. All letters are used but first letter is a vowel?
Answer
  1. Total number of $4$ letter words formed from the letters of the word $\text{'MONDAY'}$ is $={^\text{6}}\text{C}_{\text{4}}\times4!=360$
  2. Total number of words formed by using all letters of the word $\text{'MONDAY'}$ is $=6!=720$
  3. There are two vowel $A$ and $O$. So, first place can be filled in $2$ ways and the remaining $5$ places can be filled in $5!$ ways.
  4. So, tatal number of words beginning with a vowel $=2\times5!=240$
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Question 72 Marks
Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.
Answer
First separate the 3 and then arrange the remaining things.
$={^\text{n-3}}\text{C}_{\text{r-3}}(\text{r}-2)!\times3!$
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Question 82 Marks
In how many ways can a student choose $5$ courses out of $9$ courses if $2$ courses are compulsory for every student?
Answer
We have,
Out of $9$ courses $2$ are compulsory. So student can choose from $7$ courses only. Also out of $5$ courses that student need to choose, $2$ are compulsory.
So they have to choose $3$ courses out of $7$ courses. This can be done $^7C_3.$
$= 35$ ways.
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Question 92 Marks
How many different products can be obtained by multiplying two or more of the numbers 3, 5, 7, 11 (without repetition)?
Answer
The we can multiplying 2 or 3 or 4 digits.
Then number of ways of multiplying 4 digits at a time $={^\text{4}}\text{C}_{4}\ ....(\text{i})$
Then number of ways of multiplying 4 digits at a time $={^\text{4}}\text{C}_{3}\ ....(\text{ii})$
Then number of ways of multiplying 4 digits at a time $={^\text{4}}\text{C}_{2}\ ....(\text{iii})$
Total number of ways
$={^\text{4}}\text{C}_{4}+{^\text{4}}\text{C}_{2}+{^\text{4}}\text{C}_{3}$
$=1+\frac{4\times3}{2}+4$
$=11$
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Question 102 Marks
How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants?
Answer
Total vowel are = 5
Total consonants = 17
Volwels formed from 5 vowels and 17 consonants by selecting 2 vowels and 3 consonants are.
$={^\text{5}}\text{C}_{\text{2}}\times{^\text{17}}\text{C}_{\text{3}}\times5!$
$= \frac{5!}{2!3!}\times\frac{17!}{3!4!}\times120$
$=\frac{5\times4}{2}\times\frac{17\times16\times15}{3\times2}\times120$
$=10\times17\times8\times5\times120$
$=400\times17\times120$
$=6800\times120$
$=816000$
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