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CUET PG MCA Previous Year Questions (PYQs)

CUET PG MCA Permutations And Combinations PYQ


CUET PG MCA PYQ
The number of 7-digit numbers whose sum of the digits equals to 10 and which is formed by using the digits 1, 2, and 3 only is:





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CUET PG MCA Previous Year PYQ CUET PG MCA CUET 2022 PYQ

Solution


CUET PG MCA PYQ
If $\frac{1}{9!}+\frac{1}{10!}=\frac{x}{11!}$ then the value of x is:





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Solution

\[ \frac{1}{9!}+\frac{1}{10!} = \frac{11\cdot10}{11!}+\frac{11}{11!} = \frac{121}{11!} \;\Rightarrow\; x=121 \] ✅ Answer: \(x=121\)

CUET PG MCA PYQ
Out of 5 consonants and 4 vowels, how many words of 3 consonants and 3 vowels can be made? 
1.40 
2.80 
3.20 
4.240





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Solution

We have 5 consonants and 4 vowels. We need to form a word with 3 consonants and 3 vowels.

Step 1: Choose consonants

\[ \binom{5}{3} = 10 \]

Step 2: Choose vowels

\[ \binom{4}{3} = 4 \]

Step 3: Arrange the chosen 6 letters

\[ 6! = 720 \]

Step 4: Total words

\[ 10 \times 4 \times 720 = 28800 \]

Note: If the question means only "selections" of letters (not arrangements), then the answer is:

\[ \binom{5}{3}\times \binom{4}{3} = 10 \times 4 = 40 \]

Final Answer:

- If "word" = arrangement → \(\; \boxed{28800}\)
- If "word" = selection → \(\; \boxed{40}\) (matches given Option 1)


CUET PG MCA PYQ
There are 15 points in a plane such that 5 points are collinear and no three of the remaining points are collinear then total number of straight lines formed are:





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Solution

Solution:

Number of straight lines from \(n\) points (no three collinear) is \(\binom{n}{2}\).

Here, \(n = 15\).

\[ \binom{15}{2} = \frac{15 \times 14}{2} = 105 \]

Adjustment for collinearity:
Out of 15 points, 5 are collinear.
Lines from these 5 points = \(\binom{5}{2} = 10\).
But actually they form only 1 line.
Extra counted = \(10 - 1 = 9\).

Correct total lines:

\[ 105 - 9 = 96 \]

Final Answer: The total number of straight lines formed = 96


CUET PG MCA PYQ
If permutaiton of the letters of the word ‘AGAIN’ are arranged in the order as in a dictionary then 49th word i





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Solution


CUET PG MCA PYQ
4 Indians, 3 Americans and 2 Britishers are to be arranged around a round table. Answer the following questions.

The number of ways arranging them is :





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Solution

Case 1: All persons are distinct

For n distinct persons around a round table (rotations same), arrangements = \((n-1)!\).

Here, \(n=9\). So arrangements = \((9-1)! = 8! = \boxed{40{,}320}\).

Case 2: Persons are identical by nationality

If 4 Indians, 3 Americans, and 2 Britishers are considered identical within their groups, then

Arrangements = \[ \frac{(n-1)!}{4!\,3!\,2!} = \frac{8!}{4!\,3!\,2!} = \boxed{140}. \]

Final Answer:
• If all 9 are distinct → \(40{,}320\) ways.
• If only nationality matters → \(140\) ways.

CUET PG MCA PYQ
4 Indians, 3 Americans and 2 Britishers are to be arranged around a round table. Answer the following questions.

The number of ways arranging them so that the two Britishers should never come together is:





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CUET PG MCA Previous Year PYQ CUET PG MCA CUET 2022 PYQ

Solution


CUET PG MCA PYQ
4 Indians, 3 Americans and 2 Britishers are to be arranged around a round table. Answer the following questions.

The number of ways of arranging them so that the three Americans should sit together is:





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CUET PG MCA Previous Year PYQ CUET PG MCA CUET 2022 PYQ

Solution

Short Solution:

Total people = 4 Indians + 3 Americans + 2 Britishers = 9
Since arrangement is around a circular table, we fix one position ⇒ remaining to arrange: 8 positions

Group the 3 Americans together as a single unit ⇒ total units = 4 Indians + 1 American group + 2 Britishers = 7 units
Circular arrangement of 7 units = \( (7 - 1)! = 6! \)

Internal arrangements of 3 Americans = \( 3! \)

Total arrangements =
$$6! \times 3! = 720 \times 6 = \boxed{4320}$$


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