JEE Main 16 March 2021 Shift 1 Solved Paper

Set S:{0,1,2,3}
Let A=[

a1	a2	a3
b1	b2	b3
c1	c2	c3

]
AT=[

a1	b1	c1
a2	b2	c2
a3	b3	c3

]
AAT=[

a⋅a	a⋅b	a⋅c
b⋅a	b⋅b	b⋅c
c⋅a	c⋅b	c⋅c

]
where a‌‌=a1

∧

i

+a2

∧

j

+a3

∧

k

b‌‌=b1

∧

i

+b2

∧

j

+b3

∧

k

c‌‌=C1

∧

i

+C2

∧

j

+c3

∧

k

Now, as per question
a⋅a+b⋅b+c⋅c=9
[ai,bi,ci∈S]
9=(1+1+1+1+1+1+1+1+1)
or ‌‌9=(1+4+4+0+0+0+0+0+0)
or 9=(9+0+0+0+0+0+0+0+0)
or ‌‌9=(4+1+1+1+1+1+0+0+0)
Total permutations in case 1=1
Total permutations in case 2=‌

9!

6!2!

=252
In case 3=‌

9!

8!

=9
In case 4=‌

9!

5!3!

=504
Total permutations =1+252+9+504
=766