Will anyone help me to solve this puzzle?
Answers
Answer:
the answer will be that 1,2 and 3
Answer:
Please mark me brainliest
Step-by-step explanation:
1) Making the codè one place longer is correct answer.
Explanation:-
Here's the concept of Permutation i.e. number of arrangements is used. Here arrangement does matter because 3 digit codè can be like ABC or BCA or CBA etc.
For selecting r letters from n number of letters, the formula is given by \sf^nP_r = \dfrac{n!}{(n - r)!}
n
P
r
=
(n−r)!
n!
where n! represents factorial of n.
If we make codés one place longer, the number of arrangements is given by,
\implies \sf^6P_4 = \dfrac{6!}{(6- 4)!}⟹
6
P
4
=
(6−4)!
6!
\implies \sf^6P_4 = \dfrac{6!}{2!}⟹
6
P
4
=
2!
6!
\implies \sf^6P_4 = \dfrac{6 \times 5 \times 4 \times 3 \times 2!}{2!}⟹
6
P
4
=
2!
6×5×4×3×2!
\implies \sf^6P_4 =6 \times 5 \times 4 \times 3⟹
6
P
4
=6×5×4×3
\boxed{\tt{ \implies ^6P_4 =360}}
⟹
6
P
4
=360
\rule{280}{1}
But, if we add 1 more letter key, number of arrangements is given by,
\implies \sf^7P_3= \dfrac{7!}{(7- 3)!}⟹
7
P
3
=
(7−3)!
7!
\implies \sf^7P_3= \dfrac{7!}{4!}⟹
7
P
3
=
4!
7!
\implies \sf^7P_3= \dfrac{7 \times 6 \times 5 \times 4!}{4!}⟹
7
P
3
=
4!
7×6×5×4!
\implies \sf^7P_3=7 \times 6 \times 5⟹
7
P
3
=7×6×5
\boxed { \tt\implies ^7P_3=210}
⟹
7
P
3
=210
Clearly there are more number of arrangements when we make codés one place longer. So the lock will be more safe in this case.
Hence Option [1] is correct.