2 Marks Questions

Question 12 Marks

Evaluate the following:

¹²C₁₀

Answer

We have,
$=\frac{12!}{10!(12-10)!}$
$=\frac{12\times11\times10!}{10!\times2\times1}$
$=66$

Question 22 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:

4 letters are used at a time.
All letters are used at a time.
All letters are used but first letter is a vowel?

Answer

Total number of 4 letter words formed from the letters of the word 'MONDAY' is $={^\text{6}}\text{C}_{\text{4}}\times4!=360$
Total number of words formed by using all letters of the word 'MONDAY' is $=6!=720$
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 32 Marks

Evaluate the following:
¹⁴C₃

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 42 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 52 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 ⁷C₃.
= 35 ways.

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Question 62 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 72 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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Question 82 Marks

There are 10 persons named P₁, P₂, P₃, ...., P₁₀. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P₁ must occur whereas P₄ and P₅ do not occur. Find the number of such possible arrangements.

Answer

Total persons = 10
Number of persons to be selected = 5
Condition = P₁ must and P₄, P₅ 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 92 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 102 Marks

Evaluate the following:
³⁵C₃₅

Answer

We have,
$=\frac{35!}{35!(35-35)!}$
$=1$

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