118911
If \(a, b, c\) are three natural numbers in \(A P\) and \(\mathbf{a}+\mathbf{b}+\mathbf{c}=21\) then the possible number of values of the ordered triplet \((a, b, c)\) is
1 15
2 14
3 13
4 None of these
Explanation:
C We have, Here, a, b, c are in A.P \(\therefore \quad 2 \mathrm{~b}=\mathrm{a}+\mathrm{c}\) Given, \(\mathrm{a}+\mathrm{b}+\mathrm{c} =21\) \(3 \mathrm{~b} =21\) \(\mathrm{~b} =7\) \(a+b+c =21\) \(a+c =14\) The total number of solution of this equation is, \(={ }^{14-1} \mathrm{C}_{2-1}\) \(={ }^{13} \mathrm{C}_1\) \(=13\)Hence, possible number of ordered triplets is 13 .
BITSAT-2020
Permutation and Combination
118912
Find the value of \({ }^1 P_1+2 .{ }^2 P_2+3 .{ }^3 P_3+4 .{ }^4 P_4+\ldots+n .{ }^n P_n\)
118911
If \(a, b, c\) are three natural numbers in \(A P\) and \(\mathbf{a}+\mathbf{b}+\mathbf{c}=21\) then the possible number of values of the ordered triplet \((a, b, c)\) is
1 15
2 14
3 13
4 None of these
Explanation:
C We have, Here, a, b, c are in A.P \(\therefore \quad 2 \mathrm{~b}=\mathrm{a}+\mathrm{c}\) Given, \(\mathrm{a}+\mathrm{b}+\mathrm{c} =21\) \(3 \mathrm{~b} =21\) \(\mathrm{~b} =7\) \(a+b+c =21\) \(a+c =14\) The total number of solution of this equation is, \(={ }^{14-1} \mathrm{C}_{2-1}\) \(={ }^{13} \mathrm{C}_1\) \(=13\)Hence, possible number of ordered triplets is 13 .
BITSAT-2020
Permutation and Combination
118912
Find the value of \({ }^1 P_1+2 .{ }^2 P_2+3 .{ }^3 P_3+4 .{ }^4 P_4+\ldots+n .{ }^n P_n\)
118911
If \(a, b, c\) are three natural numbers in \(A P\) and \(\mathbf{a}+\mathbf{b}+\mathbf{c}=21\) then the possible number of values of the ordered triplet \((a, b, c)\) is
1 15
2 14
3 13
4 None of these
Explanation:
C We have, Here, a, b, c are in A.P \(\therefore \quad 2 \mathrm{~b}=\mathrm{a}+\mathrm{c}\) Given, \(\mathrm{a}+\mathrm{b}+\mathrm{c} =21\) \(3 \mathrm{~b} =21\) \(\mathrm{~b} =7\) \(a+b+c =21\) \(a+c =14\) The total number of solution of this equation is, \(={ }^{14-1} \mathrm{C}_{2-1}\) \(={ }^{13} \mathrm{C}_1\) \(=13\)Hence, possible number of ordered triplets is 13 .
BITSAT-2020
Permutation and Combination
118912
Find the value of \({ }^1 P_1+2 .{ }^2 P_2+3 .{ }^3 P_3+4 .{ }^4 P_4+\ldots+n .{ }^n P_n\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Permutation and Combination
118911
If \(a, b, c\) are three natural numbers in \(A P\) and \(\mathbf{a}+\mathbf{b}+\mathbf{c}=21\) then the possible number of values of the ordered triplet \((a, b, c)\) is
1 15
2 14
3 13
4 None of these
Explanation:
C We have, Here, a, b, c are in A.P \(\therefore \quad 2 \mathrm{~b}=\mathrm{a}+\mathrm{c}\) Given, \(\mathrm{a}+\mathrm{b}+\mathrm{c} =21\) \(3 \mathrm{~b} =21\) \(\mathrm{~b} =7\) \(a+b+c =21\) \(a+c =14\) The total number of solution of this equation is, \(={ }^{14-1} \mathrm{C}_{2-1}\) \(={ }^{13} \mathrm{C}_1\) \(=13\)Hence, possible number of ordered triplets is 13 .
BITSAT-2020
Permutation and Combination
118912
Find the value of \({ }^1 P_1+2 .{ }^2 P_2+3 .{ }^3 P_3+4 .{ }^4 P_4+\ldots+n .{ }^n P_n\)