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Maths Class 12th Permutations and Combinations Maths NCERT Exemplar Solutions Class 12th Chapter Seven

Numbers are to be formed between 1000 and 3000, which are divisible by 4, using the digits 1,2,3,4,5 and 6 without repetition of digits. Then the total number of such numbers is………

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alok kumar singh | Contributor-Level 10

2 months ago

Last two digit must be in form

$\begin{array}{l} 2 3, 4, 5 1 6 \\ 3 2 4 \\ 3 2 \\ 5 2 \end{array}} 3 \times 4 = 12$

Total number of required number = 12 + 18 = 30

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Let b₁ b₂ b₃ b₄be a 4-element permutation with $b_{i} \in {1, 2, 3, . . . ., 100} f o r 1 \leq i \leq 4 a n d b_{i} \neq b_{j}$ for $i \neq j,$ such that either b_1, b₂, b₃ are consecutive integers or b₂, b₃, b₄ are consecutive integers. Then the number of such permutations b₁ b₂b₃ b₄ is equal to……………

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

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A class contains b boys and g girls. If the number of ways of selecting 3 boys and 2 girls from the class is 168, then b + 3g is equal to…………..

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Kindly consider the following figure

The total number of 5-digit numbers, formed by using the digits 1, 2, 3, 5, 6, 7 without repetition, which are multiple of 6, is

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Sum of digits

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

So, either 3 or 6 rejected at a time

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……….2

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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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$A = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}], a_{2}, b_{2}, c_{2} \in {0, 1}$

S = $a_{1} + a_{2} + a_{3} + b_{1} + b_{2} + b_{3} + c_{1} + c_{2} + c_{3}$ is prime

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We have, $1 -$ (probability of all shots result in failure) $> \frac{1}{4}$

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Numbers are to be formed between 1000 and 3000, which are divisible by 4, using the digits 1,2,3,4,5 and 6 without repetition of digits. Then the total number of such numbers is………

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