AMU MCA Permutations And Combinations Previous Year Questions (PYQs)
A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations
Study Stuff for MCA Examinations
Showing 7 questions
A five digits number is formed by writing the digits $1,2,3,4,5$ in a random order without repetitions. Then the probability that the number is divisible by $4$ is
1
2
3
4
Solution
The number of ways in which $9$ persons can be divided into three equal groups is
1
2
3
4
Solution
A dictionary is printed consisting of $7$ letters words only that can be made with letters of word $CRICKET$. If the words are printed at the alphabetical order as in an ordinary dictionary, then the number of words before the word $CRICKET$ is
1
2
3
4
Solution
A five-digit number divisible by $3$ is to be formed using the numbers $0,1,3,4$ and $5$ without repetition. The total number of ways this can be done is
1
2
3
4
Solution
The number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, then the $50$th word will be
1
2
3
4
Solution
Letters: A, A, G, I, N
Words starting with A:
$ \frac{4!}{2!}=12 $
Starting with G:
$ \frac{4!}{2!}=12 $
Starting with I:
$ \frac{4!}{2!}=12 $
Total till I: $12+12+12=36$
Next start with N
Remaining: A, A, G, I
Arrangements:
$ \frac{4!}{2!}=12 $
So positions $37$ to $48$
$49^{th}$ word = NAAGI
$50^{th}$ word = NAAIG
If $^nP_4 = 20 \times ^nP_2$. Then, the value of $n$ is:
1
2
3
4
Solution
Given, $^nP_4 = 20 \times ^nP_2$$\frac{n!}{(n - 4)!} = 20 \times \frac{n!}{(n - 2)!}$$\frac{1}{(n - 4)!} = \frac{20}{(n - 2)(n - 3)(n - 4)!}$$(n - 2)(n - 3) = 20$$n^2 - 3n - 2n + 6 = 20$$n^2 - 5n + 6 - 20 = 0$$n^2 - 5n - 14 = 0$$(n - 7)(n + 2) = 0$This gives $n = 7$ or $n = -2$.Since $n$ must be a positive integer ($n \geq 4$ for $^nP_4$), $n = -2$ is not possible.$\therefore n = 7$
If $\frac{^nP_5}{^nP_3} = 20$, then the value of $n$ is:
1
2
3
4
Solution
$\frac{^nP_5}{^nP_3} = \frac{n!/(n-5)!}{n!/(n-3)!}$
$= \frac{(n-3)!}{(n-5)!}$
$= (n-3)(n-4)$
So,
$(n-3)(n-4) = 20$
$n^2 - 7n + 12 = 20$
$n^2 - 7n - 8 = 0$
$(n-8)(n+1) = 0$
$n = 8$
AMU MCA
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
AMU MCA
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
View More
Ask Your Question or Put Your Review.
