(B) $3! = 1 \times 2 \times 3 = 6$$4! = 1 \times 2 \times 3 \times 4 = 24$$\therefore 3! + 4! = 6 + 24 = 30$$7! = 1 \times 2 \times 3 \times 4 \times

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(B) $3! = 1 \times 2 \times 3 = 6$$4! = 1 \times 2 \times 3 \times 4 = 24$$\therefore 3! + 4! = 6 + 24 = 30$$7! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 = 5040$$\therefore 3! + 4! \neq 7!$

Class 11 Mathematics Permutation and Combination Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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Is $3! + 4! = 7!$?

A
Yes
B
No
C
Cannot be determined
D
None of these

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Permutation and Combination

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

How many words can be formed using any $3$ letters from the word $UNIVERSAL$?

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How many numbers greater than $1000$ but not greater than $4000$ can be formed using the digits $0, 1, 2, 3, 4$ if repetition of digits is allowed?

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The number of $4$-letter words, with or without meaning, each consisting of two vowels and two consonants that can be formed from the letters of the word $INCONSEQUENTIAL$, without repeating any letter, is:

DifficultJEE MAIN 2026

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The number of distinct positive integers that can be formed using $0, 1, 2, 3$,where each integer is used at most once,is equal to

EasyAP EAMCET 2021

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$(n - r + 1) \times ^nP_{r - 1} = \dots$

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Is $3! + 4! = 7!$?

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How many words can be formed using any $3$ letters from the word $UNIVERSAL$?

How many numbers greater than $1000$ but not greater than $4000$ can be formed using the digits $0, 1, 2, 3, 4$ if repetition of digits is allowed?

The number of $4$-letter words, with or without meaning, each consisting of two vowels and two consonants that can be formed from the letters of the word $INCONSEQUENTIAL$, without repeating any letter, is:

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