This set contains counting concept questions with solutions — Set 1 (Q1-Q10) covering a mix of question types and difficulty levels — from basic to advanced — exactly as asked in real competitive exams.
Solutions are written in a simple, step-by-step notebook style for easy self-study and quick understanding. Each solution is broken down step by step so even the toughest question feels easy. These questions are hand-picked for students preparing for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and all campus placement aptitude tests. International students preparing for GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests will find these equally useful.
✏️ Attempt each question on your own first — then check the solution below.
Counting Questions 1 to 10 with Solutions
1. if a! + b! + c! + d! + e! +f! + g! + h! is a two digit number. Find the maximum value of a.
2. If unit digit of x! + y! + z! is 9 then find the value of x!.y!.z!.
3. Find the unit digit of 1! + 2! + 3! + 4! + 5! +6! + ………… +15!
4. If N = p! + q! + r! is a two digit prime number then how many such value of n exists?
5. If product of factorial of n consecutive integer is a two digit number then find the maximum value of n.
6. If sum of factorial of n consecutive natural nmbers ia a four digit number then find the maximum value of n.
7. Find highest power of 12 in 1212!
8. Find the highest power of 20 in 10!×20!×30!×40!×50!×60!×70!×80!×90!×100!×110!×120!
9. If N is product of factorial of four consecutive positive integers. Then which of the following is/are correct about N.
(i) if N is a three digit number then there exists two such value of N.
(ii)if N is a Five digit number then there exists only one such value of N
(iii) N may be a four digit number
10. Find the highest power of 10 in 10! + 20! + 30! + 40! + 50! + 60! + 70! + 80! + 90! + 100! + 110! + 120!
Counting Questions 1 to 10 — Step-by-Step Solutions
1. If a! + b! + c! +d! +e! +f! +g! +h! is a two digit number then find the maximum value of a.
Solution:- For maximum value of a → value of other factorial numbers should be minimum (i.e. 1)
Hence b! + c! + d! + e! + f! + g! + h! = 7
1 1 1 1 1 1 1
& Since given expression is a two digit number. Hence maximum value of a! should be less than 99 – 7 = 92
a! ≤ 92
& Since 4! = 24
& 5! = 120
Hence maximum value of a = 4 Answer
2. If unit digit of x! + y! + z! is 9 then find the value of x!.y!.z!.
Sol:-
0! = 1
1! = 1
2! = 2
3! = 6
4!= 24
5! = 120
6! = 720
Hence all factorials greater than or equal to 5 will have 0 at their unit place. so they are not of our use in this question.
in both the cases:-
x!.y!.z! = 0!.2!.3! = 1!.2!.3! = 12 Answer
3. Find the unit digit of 1! + 2! + 3! + 4! + 5! + 6! + ………. + 14! + 15!
Sol:-
4. If N = p! + q! + r! is a two digit prime number. Then how many such value of N exist.
Solution:-
∴ 0! + 3! + 4! = 31 & 1! + 3! + 4! = 31
↓ ↓
prime prime
& no other cases
Hence 2 such values of N exists. Answer
5. If product of factorials of n consecutive integers is a two digit number then find the maximum value of n.
Solution:-
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120 →
For maximum value of n ⇨
Hence max(n) = 4 Answer
6. If sum of factorial of n consecutive natural numbers is a four digit number then find the maximum value of n.
Solution:-
Hence max(n) = 7 Answer
7. find the highest power of 12 in 1212!.
Solution:- 12 = 2² × 3
∴ Highest power of 12 in 1212! = minimum of above two values = 601 Answer
8. Find the highest power of 20 in 10! × 20! × 30! × 40! × 50! × 60! × 70! × 80! × 90! × 100! × 110! × 120!.
Solution:-
20 = 2² × 5
for 10! ⇒
Since out of given factorial numbers 10! is least & in 10! highest power of 5 (i.e. 2) is less than highest power of 2² (i.e. 4)
Hence as the value of factorial number ↑, highest power of 2 or 2² will always be greater than highest power of 5.
So for maximum power of 20 we have to find only max power of 5 in every factorial & add all of them.⇒
9. If N is a product of factorial of four consecutive positive integers then which of the following is/are correct about N.
(i) if N is a three digit number then there exists two such value of N. ➡ ✖
(ii) if N is a Five digit number then there exists only one such value of N. ➡ ✔
(iii) N may be a four digit number. ➡✖
Solution:-
if N = 1!.2!.3!.4! = 288
N = 2!.3!.4!.5! = 34560
N = 3!.4!.5!.6! = 12441600
N = 4!.5!.6!.7! = 10450944000
10. Find the highest power of 10 in 10! + 20! + 30! + 40! + 50! + 60! + 70! +80! + 90! + 100! + 110! + 120!.
Solution:-
From earlier question we know that highest power of 10 in multiplication of theser factorials is 182. Now herefactorials are written in summation form. Hence we have to find the highest power of 10 or in other words number of trailing zeros in the factorial summation.
As the value of factorial number increases, the number of trailing zeros at its end will also increases. Hence out of these 10! which has least number of trailing zeros at its end & all other factorials will have more number of trailing zeros than 10!
Hence highest power of 10 = number of trailing zeros in the summation = highest power of 10 in 10! ➡
✅ Well done on completing Set 1!
Continue practising with Counting Concept Questions 11 to 20 → Set 2 or revisit the Counting Concept Page to strengthen your formulas and tricks before moving ahead.
Consistent practice is the key to mastering counting concept for SSC CGL, SSC CHSL, CAT, Bank PO, Bank Clerk, UPSC CSAT, Railway RRB, AMCAT, eLitmus, TCS NQT and international exams including GRE, GMAT, SAT, ACT, MAT and all Numerical Reasoning Tests. Want to understand the concept better? Read about Factorial on Wikipedia before attempting the next set.
This page is part of our complete series of counting concept questions with solutions for competitive exams — covering every question type from basic to advanced so you can build speed, accuracy and confidence. Practising these questions regularly will also strengthen your core counting concept before your exam day.