NIMCET 2027 Permutation And Combination Previous Year Questions (PYQs) – Page 1 of 2
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Showing 14 questions
How many 3-digit numbers divisible by 5, can be formed using the digits 2 3 5 6 7 and 9, without
repetition of digits?
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4
Solution
A number divisible by 5 must end in 5.
Fix the units digit as 5.
Remaining digits for hundreds and tens places: 2, 3, 6, 7, 9 (5 digits).
Number of ways to choose hundreds and tens digits = 5 × 4 = 20.
Answer: 20
Using the digits 1,5,2,8 all possible four digit numbers are formed and the sum of all such numbers is between
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4
Solution
Digits = 1,5,2,8
Number of arrangements = 4! = 24
Each digit appears 6 times in each place (unit, tens, hundreds, thousands).
Sum contributed by one digit = 6 × (1000 + 100 + 10 + 1) × digit
= 6 × 1111 × digit = 6666 × digit
Sum of digits = 1 + 5 + 2 + 8 = 16
Total sum = 6666 × 16 = 106656
This lies between 50000 and 100000.
All the letters of the word 'INDIA' are permuted… the 58th word in dictionary order is?
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Solution
Word: INDIA
Letters in alphabetical order: A, D, I, I, N (5 letters)
Number of permutations starting with each leading letter:
● Starting with A:
Remaining: D, I, I, N → total = 4! / 2! = 12 words
● Starting with D:
Remaining: A, I, I, N → again 12 words
Cumulative: 24 words
● Starting with I (first I):
Remaining: A, D, I, N → 4 letters all different = 24 words
Cumulative: 48 words
Next block starts with I again (second I):
We need the 58th word → 58 − 48 = 10th word in this block.
Now list 4-letter permutations of A, D, I, N:
Alphabetical order: A, D, I, N
Block starting with I gives words:
I A D N
I A N D
I A I N
I A N I
I D A N
I D N A
I D I A
I I A D
I I A N
I I D A → This is the 10th.
So final word = NIIDA? No, because leading letter should be I?
Wait: The initial letter of the block is N, not I — Let's correct.
We already used A (12), D (12), I (24).
Next letter in order is N.
So 48 words before N.
We need 58 → 10th under “N…”.
Remaining letters: A, D, I, I
Sorted: A, D, I, I
10th permutation of A, D, I, I is IDIA → attach N at start:
Correct word = NIDIA
At a dance party a group of girls and boys exchange dances as follows:
One boy dances with 5 girls. Second boy dances with 6 girls, and so on; last boy dances with all girls. If b represents the number of boys and g represents the number of girls, then
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Solution
Number of girls danced with by boys:
1st boy → 5 girls
2nd boy → 6 girls
3rd boy → 7 girls
…
last boy → g girls
This is an arithmetic sequence: 5, 6, 7, …, g
Number of terms = b = (last – first) + 1 = (g – 5) + 1 = g – 4
Ten points are marked on a straight line and eleven points are marked on another straight line. How many triangles can be constructed with vertices from among the above points?
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Solution
We can get the triangles in
two different ways.
Taking two points from the line having 10 points
(in
So,
the number of triangles here is
Taking
two points from the line having 11 points (in
Total number of triangles, =495+550
How many 4 - digit numbers that can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
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Solution
In an objective type examination, 120 objective type questions are there; each with 4 options P, Q, R and S.
A candidate can choose either one of these options or can leave the question unanswered.
How many different ways exist for answering this question paper?
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Solution
$\text{Each question has 5 choices (4 options + leave blank).}$
$\text{Total number of ways } = 5 \times 5 \times \cdots \times 5 \text{ (120 times)}$
$= 5^{120}$
How many positive numbers less than 10,000 are such that the product of their digits is 210?
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Solution
210=1*2*3*5*7=1*6*5*7.
(Only 2*3 makes the single digit 6).
So, four digit numbers with combinations of the digits {1,6,5,7} and {2,3,5,7} and three digit numbers with combinations of digits {6,5,7} will have the product of their digits equal to 210.
{1,6,5,7} # of combinations 4!=24
{2,3,5,7} # of combinations 4!=24
{6,5,7} # of combinations 3!=6
24+24+6=54.
12 members were present at a board meeting. Each member shook hands with all of other members before and after the meeting. How many handshakes were there?
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Solution
One round handshakes $=\binom{12}{2}=66$
Before + After $=2\times 66=132$
Let n be the number of different 5 digit numbers, divisible by 4 that can be formed with the digits 1,2, 3, 4, 5 and 6, with no digit being repeated. What is the value of n ?
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Solution
We have to make 5 digit no. And also divisible by 4 using digits (1,2,3,4,5,6) To be divisible by 4 last two digit of 5 digits no should be divisible by 4 so we choose last two digits st they are divisible by 4 and rest with remaining four digits. Last two digits only can be 12,16,24,32,36,52,56,64 as all are divisible by 4. So total no of ways= no ways of choosing first 3 digits* no ways of choosing last two digits Total= (4*3*2)*(8) =192
Abhijeet Kushwaha
NIMCET 2027
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NIMCET 2027
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and More.
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