Question

Let $m,n \in N$ and $\gcd \left( {2,n} \right) = 1$. If $30{}^{30}{C_0} + 29{}^{30}{C_1} + ..... + 2{}^{30}{C_{28}} + 1{}^{30}{C_{29}} = n{.2^m}$, then $n + m$ is equal to

Answer 1

Hint: In this question, given that $\gcd \left( {2,n} \right) = 1$ which implies that $n$ is an odd number. Also, given $30{}^{30}{C_0} + 29{}^{30}{C_1} + ..... + 2{}^{30}{C_{28}} + 1{}^{30}{C_{29}} = n{.2^m}$ make this series a sequence separate both the terms apply combination formula ${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ...... + {}^n{C_n} = {2^n}$. In last, compare both the sides to know the value of $m,n$ and put the required values in $n + m$.

Formula used:
Combination formula – A combination is a mathematical technique for determining the number of potential arrangements in a set of objects where the order of the selection is irrelevant. You can choose the components in any order in combinations. Combinations are studied in combinatorics, but they are also applied in other fields such as mathematics and finance.
${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ...... + {}^n{C_n} = {2^n}$

Complete step by step solution:
Given that,
$30{}^{30}{C_0} + 29{}^{30}{C_1} + ..... + 2{}^{30}{C_{28}} + 1{}^{30}{C_{29}} = n{.2^m}$
Also written as, $\sum\limits_{a = 0}^{30} {\left( {30 - a} \right){}^{30}{C_a}} = n{.2^m}$
$30\sum\limits_{a = 0}^{30} {{}^{30}{C_a} - \sum\limits_{a = 0}^{30} a {}^{30}{C_a} = n{{.2}^m}} $
Applying the combination formula i.e., ${}^n{C_0} + {}^n{C_1} + {}^n{C_2} + ...... + {}^n{C_n} = {2^n}$
We get, $30\left( {{2^{30}}} \right) - \sum\limits_{a = 1}^{30} {a\left( {\dfrac{{30}}{a}} \right)} {}^{29}{C_{a - 1}} = n{.2^m}$
\[30\left( {{2^{30}}} \right) - 30\left( {{2^{29}}} \right) = n{.2^m}\]
\[30 \times {2^{29}}\left( {2 - 1} \right) = n{.2^m}\]
On solving, we get \[15 \times {2^{30}} = n{.2^m}\]
Comparing both the sides
It implies that, \[n = 15,m = 30\]
Hence, the value of n + m = 15 + 30 = 45

Note: To solve this problem students must know how to convert a series in sequence. Formulas should be on tip. As in this question, the second term limit is changed from $0$ to $1$ because at zero we were getting the whole term zero. So, we multiplied the term with $\left( {\dfrac{{30}}{a}} \right)$ and changed the limit. Also, combination is a sort of permutation in which the order of the selection is ignored. As a result, the number of permutations is always more than the number of combinations. This is the fundamental distinction between permutation and combination.