The number of matrices of order 3 × 3, whose entries are either 0 or 1 and the sum of all the entries is a prime number, is……….

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    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago

     A= [a1a2a3b1b2b3c1c2c3], a2, b2, c2 {0, 1}

    S = a1+a2+a3+b1+b2+b3+c1+c2+c3 is prime

    0s9

    Prime value = 2,3, 5, 7

    ForS=2, 1+1+0+0+0+0+0+0+09!2!7!=36

    Total number of matrices

    = 36 + 84 + 126 + 36 = 282

Similar Questions for you

A
alok kumar singh

Three consecutive integers belong to 98 sets and four consecutive integers belongs to 97 sets.

Þ Number of permutations of b1 b2 b3 b4 = number of permutations when b1 b2 b3 are consecutive + number of permutations when b2, b3, b4 are consecutive – b1 b2 b3 b4 are consecutive = 98 * 97 * 98 * 97 – 97 = 18915

A
alok kumar singh

Last two digit must be in form

23, 4, 5163243252}3×4=12

Total number of required number = 12 + 18 = 30

V
Vishal Baghel

Kindly consider the following figure

P
Payal Gupta

Sum of digits

1 + 2 + 3 + 5 + 6 + 7 = 24

So, either 3 or 6 rejected at a time

Case 1 Last digit is 2

……….2

no. of cases = 2C1 × 4! = 48

Case 2 Last digit is 6

……….6

= 4! = 24

Total cases = 72

A
alok kumar singh

We have,  1-  (probability of all shots result in failure)  > 1 4

1 - 9 10 n > 1 4 3 4 > 9 10 n n 3

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