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MAH CET MCA Previous Year Questions (PYQs)

MAH CET MCA Permutations And Combinations PYQ

MAH CET MCA PYQ
Let $K$ be the set of all odd numbers less than $200$ and $M$ be the set of products of two distinct numbers taken from $K$, then find the total number which is divisible by $5$ in $M$.





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MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2023 PYQ

Solution

Among $100$ odd numbers, $20$ are multiples of $5$. Required count $= 100 \times 20 - C(20,2) = 392$.
MAH CET MCA PYQ
The number of diagonals in a polygon of $m$ sides is





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MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2023 PYQ

Solution

Formula for number of diagonals $= \dfrac{m(m-3)}{2}$.
MAH CET MCA PYQ
A five digit number divisible by 3 using digits 0, 1, 2, 3, 4, 5 is to be made without repetition. Find of such word.





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MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2023 PYQ

Solution

Digits available: {0,1,2,3,4,5}. 

Sum of all six = 15 divisible by 3.
To make a 5-digit number divisible by 3, exclude one digit whose value 0, i.e., exclude 0 or 3.

  • Exclude 0 → use {1,2,3,4,5}: all 5! = 120 numbers (no leading-zero issue).

  • Exclude 3 → use {0,1,2,4,5}: total 5! − 4! (leading 0) = 120 − 24 = 96.

Total = 120 + 96 = 216

MAH CET MCA PYQ
Find the total number of ways to put $20$ balls into $5$ boxes so that the first box contains only one ball.





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MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2025 (Shift 1) PYQ

Solution

Choose which ball goes to Box 1: $20$ ways. Each of the remaining $19$ balls can go to any of Boxes $2$–$5$: $4^{19}$ ways. Total $=20\cdot4^{19}$.
MAH CET MCA PYQ
Find the number of arrangements of the letters of the word $,\text{INDEPENDENCE},$ such that
(i) the words start with $I$ and end with $P$.
(ii) all the vowels never occur together.





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MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2025 (Shift 1) PYQ

Solution

Letters in INDEPENDENCE: $E^4,\ N^3,\ D^2,\ I,\ P,\ C$ (total $12$).

(i) Fix $I$ at start and $P$ at end. Arrange the remaining $10$ symbols: $E^4,\ N^3,\ D^2,\ C$.
Number $= \dfrac{10!}{4!,3!,2!}=12600$.

(ii) Total arrangements $= \dfrac{12!}{4!,3!,2!}=1663200$.
“All vowels together”: treat $(\text{vowels}=E^4I)$ as one block. Then items are
${\text{block}, N^3, D^2, C, P}$ → $8$ items with repeats $N^3, D^2$.
Arrangements of items $= \dfrac{8!}{3!,2!}=3360$.
Internal vowel-block arrangements $= \dfrac{5!}{4!}=5$.
Together $= 3360\times5=16800$.
Hence “never together” $=1663200-16800=1646400$.
MAH CET MCA PYQ
Straight lines are drawn by joining m points on a straight line to n points on another line. Then excluding the given points, the number of point of intersection of the lines drawn is (no two lines drawn are parallel and no three lines are concurrent).





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MAH CET MCA Previous Year PYQ MAH CET MCA MAH MCA CET 2024 PYQ

Solution


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