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Binomial Theorem · Gujarat Board (GSEB) STD 11 Science (GSEB - 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 $P _1, P _2, P _3, \ldots, P _{10}$. 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 $10-3=7={ }^7 C _4 \times 5$ !
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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 '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 'MONDAY' is $={^\text{6}}\text{C}_{\text{4}}\times4!=360$
  2. Total number of words formed by using all letters of the word '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.
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 ${ }^7 C_3$. $=35$ 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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