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Question: Find the number of ways in which 5-boys and 5-girls be seated in a row so that all the girls sit tog...

Find the number of ways in which 5-boys and 5-girls be seated in a row so that all the girls sit together and all the boys sit together.

Explanation

Solution

Hint: We will be using the concept of permutation and combination in this question. We know that the number of permutation of r things among n things is npr^{n}{{p}_{r}} and the number of combinations of r things among n things is ncr^{n}{{c}_{r}}.
Complete step by step answer:
Now to solve the question we have to find the number of total possible ways in which 5-boys and 5-girls can be seated in a row so that the 5-boys and 5-girls sit together.
We know that the number of boys =5.
Also, the number of girls =5.
We will now permute by assuming all the 5 boys as one and all the 5 girls as one because we have to make sure that all the boys and all the girls sit together. So, the number of ways in which it can be done is 2p2^{2}{{p}_{2}}= 2!
Now we have the number of ways in which all the 5 boys as a whole and all the 5 girls as a whole can sit together but the girls and boys can also be permuted in among themselves in 5p5=5!^{5}{{p}_{5}}=5!.
Now we have to find the total possible combinations of permuting 5-boys and 5-girls by using the multiplication rule of permutation which can be stated simply as the idea that if there are ways of doing something and b ways of doing another thing, then there are a×ba\times b ways of performing both actions this will be equal to 2!×\times 5! ×\times 5! = 28,800.

Note: In these types of questions there is a high chance that we miss some possible combination so it was always advisable to recheck the approach once again. Also, some common formulas like npr^{n}{{p}_{r}}=n!(n−r)!\dfrac{n!}{(n-r)!} and ncr^{n}{{c}_{r}}=n!r!(n−r)!\dfrac{n!}{r!(n-r)!} are required to solve the problem so memorizing these formulae shortens the calculation.