^{14}C_4 + sum_{j=1}^4 {^{18-j}C_3} is equal | vedclass

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(B) The given expression is $^{14}C_4 + \sum_{j=1}^4 {^{18-j}C_3}$.Expanding the summation,we get: $^{14}C_4 + (^{17}C_3 + ^{16}C_3 + ^{15}C_3 + ^{14}

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$^{18}C_4$

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(B) The given expression is $^{14}C_4 + \sum_{j=1}^4 {^{18-j}C_3}$.Expanding the summation,we get: $^{14}C_4 + (^{17}C_3 + ^{16}C_3 + ^{15}C_3 + ^{14}C_3)$.Rearranging the terms: $(^{14}C_4 + ^{14}C_3) + ^{15}C_3 + ^{16}C_3 + ^{17}C_3$.Using the Pascal's identity $^{n}C_r + ^{n}C_{r-1} = ^{n+1}C_r$,we have $^{14}C_4 + ^{14}C_3 = ^{15}C_4$.Now,the expression becomes: $(^{15}C_4 + ^{15}C_3) + ^{16}C_3 + ^{17}C_3$.Again,using the identity: $^{15}C_4 + ^{15}C_3 = ^{16}C_4$.So,the expression becomes: $(^{16}C_4 + ^{16}C_3) + ^{17}C_3$.Again,using the identity: $^{16}C_4 + ^{16}C_3 = ^{17}C_4$.Finally,the expression becomes: $^{17}C_4 + ^{17}C_3 = ^{18}C_4$.

$^{14}C_4 + \sum_{j=1}^4 {^{18-j}C_3}$ is equal to

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