Tool By Zeal Point

Factorial : Examples 1

Q. How many 4's are there in 80! ? (প্ৰশ্ন: 80!-ত মুঠ কিমানটা 4 আছে ?)

Solⁿ:

Divide repeatedly by 4: 80 ÷ 4 = 20 , 20 ÷ 4 = 5 , 5 ÷ 4 = 1

Now add: 20 + 5 + 1 = 26

Ans: C. 26

Note: Divide by 4 until the quotient becomes less than 4

Practice Questions

How many 4's are there in 60! ?

How many 4's are there in 120! ?

How many 4's are there in 150! ?

How many 4's are there in 75! ?

How many 4's are there in 90! ?

============================================

Q. Trailing zeros in 100! ? (১০০! (১০০ ফেক্টৰিয়েল)-ত শেষত কিমানটা শূন্য (Trailing Zeros) থাকে ?

Options / বিকল্পসমূহ: A) 20 B) 25 C) 24

Trick (Trailing Zeros in )

100!

100 ÷ 5 = 20

20 ÷ 5 = 4

4 ÷ 5 = 0

Add the quotients: 20 + 4 = 24

Ans: C) 24

Rule: Keep dividing by 5 until the answer becomes 0. Add all the quotients.

Trailing Zeros in Factorial – Practice Set

Trailing zeros in 25! ? (২৫! ত শেষত কিমানটা শূন্য ?) A) 4 B) 5 C) 6 D)

Trailing zeros in 50!? (৫০! ত শেষত কিমানটা শূন্য ?) A) 10 B) 11 C) 12 D) 13

Trailing zeros in 75!? (৭৫! ত শেষত কিমানটা শূন্য ?) A) 15 B) 16 C) 17 D) 18

Trailing zeros in 125!? (১২৫! ত শেষত কিমানটা শূন্য ?) A) 30 B) 31 C) 32 D) 33

Trailing zeros in 200!? (২০০! ত শেষত কিমানটা শূন্য ?) A) 48 B) 49 C) 50 D) 52

==============================================

Q. In how many different ways can the letters of the word "MODI" be arranged ? প্ৰশ্ন: "MODI" শব্দটোৰ আখৰসমূহক কিমান ধৰণে সজাব পাৰি ?

A) 32 B) 24 C) 48 D) 64

Solution | সমাধান

The word MODI has 4 letters. ("MODI" শব্দটোত ৪টা আখৰ আছে।)

M, O, D, I

All letters are different. (সকলো আখৰ বেলেগ-বেলেগ।)

No letter is repeated. (এটা আখৰো পুনৰাবৃত্তি হোৱা নাই।)

Formula | সূত্ৰ

If all letters are different: যদি সকলো আখৰ বেলেগ হয়:

Ans | উত্তৰ: B) 24

Trick | ট্ৰিক

Number of different letters = 4 (মুঠ আখৰ = ৪টা)

All letters are different → 4! = 24 (সকলো আখৰ বেলেগ → 4! = 24)

Examples | উদাহৰণ

MODI →

BIHAR →

BOOK →

Q. How many fives are there in $125!$ ?

Options: A) 25 B) 31 C) 35 D) 40

Trick

$125 \div 5 = 25, 25 \div 5 = 5, 5 \div 5 = 1$

Add: $25 + 5 + 1 = 31$

Ans: B) 31

=================================

Q. How many factors of 2 are there in $64!$ ?

A) 55 B) 63 C) 48 D) 32

Trick

$64 \div 2 = 32, 32 \div 2 = 16,$ $16 \div 2 = 8,$ $8 \div 2 = 4, 4 \div 2 = 2,$ $2 \div 2 = 1$

Now add: $32 + 16 + 8 + 4 + 2 + 1 = 63$

Ans: B) 63

====================================

Q. How many 3s are there in $81!$ ?

Options: A) 20 B) 30 C) 40 D) 27

Trick

81 \div 3 = 27 ,

81 \div 9 = 9 ,

81 \div 27 = 3

27 + 9 + 3 = 39

Ans: 39 Since 39 is not given in the options, the nearest option is:C) 40

====================================