NIMCET Permutation and Combination Previous Year Question Paper and Solution with PDF

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NIMCET Previous Year Questions (PYQs)

NIMCET Permutation And Combination PYQ

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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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ

Solution

We can get the triangles in two different ways.

Taking two points from the line having 10 points

(in ¹⁰C2 ways, i.e., 45 ways) and one point from the line consisting of 11 points (in 11 ways).

So, the number of triangles here is 45×11=495.

Taking two points from the line having 11 points (in ¹¹C2, i.e., 55 ways) and one point from the line consisting of 10 points (in 10 ways), the number of triangles here is 55×10=550

Total number of triangles, =495+550

=1,045 triangles

Qus : 2NIMCET PYQ

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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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ

Solution

Qus : 3NIMCET PYQ

How many positive numbers less than 10,000 are such that the product of their digits is 210?

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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ

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.

Qus : 4NIMCET PYQ

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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NIMCET Previous Year PYQNIMCET NIMCET 2018 PYQ

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

Qus : 5NIMCET PYQ

What is the largest number of positive integers to be picked up randomly so that the sum of difference of any two of the chosen numbers is divisible by 10?

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NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ