This set contains counting questions with solutions — Set 3 (Q21-Q30) 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 21 to 30 with Solutions
21. Find the first two non-zero digit of (555!).(444!).
22. Find the number of trailing zeros in (5²⁰)!.
23. Find the number of trailing zeros in 1!.2!.3!.4!.5!………..148!.149!.150!.
24. If S is the sum of factorials of all the prime numbers less than 150. Then what will be the last three digit of S.
25. If x(y!) is completely divisible by 5¹⁵ & x is a single digit number. then find the minimum & maximum value of y.
26. Let p be a two digit prime number greater than 60 & k be a number such that k = \(\frac{{400! \times 300!}}{{500!}}\). Given that exponent of p in k is 2. then how many such value of p exists.
27. Find the highest power of 12 in 10² × 11² ×12² × 13² ×……….× 30² × 31².
28. Find the number of trailing zeros at the end of 40×41×42×43×………till 40 terms.
29. Find the highest power of 1000 in 1000×1001×1002×1003×………..till 1000 terms.
30. Find the first two non zero digit i.e. FTNZD in 100×101×102×………..till 100 terms.
(i) 34
(ii) 39
(iii) 24
(iv) 92
Counting Questions 21 to 30 — Step-by-Step Solutions
21. Find the first two non-zero digit FTNZD of (555!).(444!).
Solution:-
if wer multiply these two factorial numbers then number of trailing zeros in the product will be the summation of trailing zeros in these two factorial numbers.
∴ FTNZD of product = FTNZD(555!) × FTNZD(444!)
= 56 × 48
= 88 Answer
22. Find the number of trailing zeros in (5²⁰)!.
Solution:-
Number of trailing zeros =
= 5¹⁹ + 5¹⁸ + 5¹⁷ + ……….. + 5 + 1
now this is a G.P. series & we know that the sum of the G.P. is =
23. Find the number of trailing zeros in 1!.2!.3!.4!.5!………148!.149!.150!.
Solution:-
1! till 4! will have same number of trailing zeros.
5! till 9! will have same number of trailing zeros.
10! till 14! will have same number of trailing zeros.
15! till 19! will have same number of trailing zeros.
.
.
.
.
.
.
95! till 99! will have same number of trailing zeros.
.
.
.
.
.
.
140! till 144! will have same number of trailing zeros.
& 145! till 149! will have same number of trailing zeros.
So divide above number in group of five-five factorial numberes which have same number of trailing zeros.
➟ (1!.2!.3!.4!)(5!.6!.7!.8!.9!)(10!.11!.12!.13!.14!)(15!.16!.17!.18!.19!)(20!.21!.22!.23!.24!)………..(95!.96!.97!.98!.99!)……..(145!.146!.147!.148!.149!)(150!).
⇨ now number of trailing zeros in first group i.e. (1!.2!.3!.4!) = 0
Hence number of trailing zeros or highest power of 5 in given question will be the sum of all theser total values = 2612 Answer
24. If S is the sum of factorial of all the prime numbers less than 150. Then what will be the last three digit of S.
Solution:- Since the last three digits of factorials greater than or equal to 15! are ‘000’. Hence we need not to consider factorial greater than or equal to 15 because summation of these will not contribute anything in last three digits.
Hence we need to find the sum of all prime numbers less than 15!.
2! ⟶ 2
3! ⟶ 6
5! ⟶ 120
7! ⟶ 5040
11! ⟶ 39916800
13! ⟶ 6007020800
25. If x(y!) is completely divisible by 5¹⁵ & x is a single digit number. then find the minimum & maximum value of y.
Solution:-
To find the minimum value of y ➪
Since x is a single digit number & we have to find the minimum value of y:
So take x as 3
now we are to find minimum value of y such that y! is completely divisible by 5¹⁴
Hence minimum value of y should be 60. Answer
Now to find maximum value of y:
take value of x as any single digit number except 5
Hence maximum value of y = 69 Answer
26. Let p be a two digit prime number greater than 60 & k be a number such that k= \(\boldsymbol{\frac{{(400!)(300!)}}{{500!}}}\) . Given that exponent of p in k is 2. then how many such value of p exists.
Solution:
Hence total 6 such values exists. Answer
27. Find the highest power of 12 in 10² × 11² ×12² × 13² ×……….× 30² × 31².
Solution:
10² × 11² ×12² × 13² ×……….× 30² × 31²
= (10 × 11 × 12 × 13 ×……….× 30 × 31)²
= \({\left( {\frac{{31!}}{{9!}}} \right)^2}\)
so we have to find the highest power of 12 in \({\left( {\frac{{31!}}{{9!}}} \right)^2}\)
now ⇒ 12 = 2² × 3
⇒
∴ Highest power of 2 in \(\frac{{31!}}{{9!}}\) = 26 – 7 = 19
Highest power of 2 in \({\left( {\frac{{31!}}{{9!}}} \right)^2}\) = (19) × 2 = 38
Highest power of 2² in \({\left( {\frac{{31!}}{{9!}}} \right)^2}\) = \(\frac{{38}}{2}\) = 19
Now highest power of 3 in
∴ Highest power of 3 in \(\left( {\frac{{31!}}{{9!}}} \right)\) = 14 – 4 = 10
Highest power of 3 in \({\left( {\frac{{31!}}{{9!}}} \right)^2}\) = 10 + 10 = 20
Hence highest power of 12 in \({\left( {\frac{{31!}}{{9!}}} \right)^2}\)
= Minimum of (Highest power of 2² in \({\left( {\frac{{31!}}{{9!}}} \right)^2}\) & Highest power of 3 in \({\left( {\frac{{31!}}{{9!}}} \right)^2}\)
= 19 Answer
28. Find the number of trailing zeros at the end of 40 × 41 × 42 × 43 × ………..till 40 terms.
Solution:-
40 × 41 × 42 × 43 × …….. × 78 × 79
= \(\frac{{79!}}{{39!}}\)
so the question has converted that we have to find the number of trailing zeros at the end of \(\frac{{79!}}{{39!}}\)
⇒ Now the number of trailing zeros at the end of
Hence number of trailing zeros in the given equation = 18 – 8 = 10 Answer
29. Find the highest power of 1000 in 1000×1001×1002×1003×………..till 1000 terms.
Solution:-
= \(\frac{{1999!}}{{999!}}\)
now 1000 = 10×10×10 = 2³×5³
So to find maximum power of 1000 we need to find maximum power of 5 first & then divide that value by 3 (due to 5³) and Integer part of that value will be the final answer.
Hence highest power of 5 in
∴ Highest power of 5 in \(\frac{{1999!}}{{999!}}\) = 496 – 246 = 250
Hence highest power of 5³ in \(\frac{{1999!}}{{999!}}\) = \(\frac{{250}}{3}\) = 83
Hence highest power of 1000 in Given question = 83 Answer
30. Find the first two non zero digit i.e. FTNZD in 100×101×102×………..till 100 terms.
Solution:-
= \(\frac{{199!}}{{99!}}\)
So we have to find FTNZD in \(\left( {\frac{{199!}}{{99!}}} \right)\)
first find FTNZD(199!) ⇒
∴ FTNZD(199!) = 12⁴⁷ × Uₜ(24×24×42×1)
= (2²×3)⁴⁷ × Uₜ(76×42)
= 2⁹⁴ × 3⁴⁷ × 92
= (2²⁰)⁴ × 2¹⁴ × (81)¹¹ ×27 ×92
= 76 × 84 × 81 × 27 × 92
= 36
Now to find FTNZD(99!)
∴ FTNZD(99!) = 12²² × Uₜ(24×24×6)
= (2²×3)²² × Uₜ(76×6)
= 2⁴⁴ × 3²² × 56
= (2²⁰)² × 2⁴ × (81)⁵ × 3² × 56
= 76 × 16 × 01 × 9 × 56
= 64
Now 199! has 47 trailing zeros & 99! has 22 trailing zeros
∴ \(\frac{{199!}}{{99!}}\) will have 47 – 22 = 25 trailing zeros
So FTNZD of \(\left( {\frac{{199!}}{{99!}}} \right)\) will be the division of FTNZD of 199! & FTNZD of 99!
∴FTNZD\(\left( {\frac{{199!}}{{99!}}} \right)\) = \(\frac{{FTNZD(199!)}}{{FTNZD(99!)}}\)= \(\frac{{36}}{{64}}\)
From given options Uₜ(64×24) = 36
Hence Answer = 24 option (iii) is correct.
✅ Well done on completing Set 3!
Continue practising with Counting Concept Questions 31 to 40 → Set 4 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.