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Question: If \[{}^{56}{{P}_{r+6}}:{}^{54}{{P}_{r+3}}=30800:1\], find r....

If 56Pr+6:54Pr+3=30800:1{}^{56}{{P}_{r+6}}:{}^{54}{{P}_{r+3}}=30800:1, find r.

Explanation

Solution

Hint: The expression is that of Permutation, which represents ordered matters. For number of permutation of n things taken r at a time = nPr=n!(n−r)!{}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}. Simplify the given expression with this formula and find the value of r.

Complete step-by-step answer:
Permutation of a set is an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its element. Permutation is also the linear order of an ordered set. Thus the number of permutation (ordered matters) of n things taken r at a time is given as,
nPr=P(n,r)n!(n−r)!{}^{n}{{P}_{r}}=P\left( n,r \right)\dfrac{n!}{\left( n-r \right)!}
Now, we have been given that,
56Pr+6:54Pr+3=30800:1…...(1){}^{56}{{P}_{r+6}}:{}^{54}{{P}_{r+3}}=30800:1…...(1)
Let us simplify it as per the formula of Permutation.
56Pr+6=56!(56−r−6)!=56!(50−r)!{}^{56}{{P}_{r+6}}=\dfrac{56!}{\left( 56-r-6 \right)!}=\dfrac{56!}{\left( 50-r \right)!}
Similarly, 54Pr+3=54!(54−r−3)!=54!(51−r)!{}^{54}{{P}_{r+3}}=\dfrac{54!}{\left( 54-r-3 \right)!}=\dfrac{54!}{\left( 51-r \right)!}
Now let us substitute the formula of 54Pr+6{}^{54}{{P}_{r+6}} and 54Pr+3{}^{54}{{P}_{r+3}} in Equation (1)
56Pr+6:54Pr+3=30800:1{}^{56}{{P}_{r+6}}:{}^{54}{{P}_{r+3}}=30800:1
56!(50−r)!:54!(51−r)!=30800:1\dfrac{56!}{\left( 50-r \right)!}:\dfrac{54!}{\left( 51-r \right)!}=30800:1, we can write the above as,

56!(50−r)!54!(51−r)!=308001\dfrac{\dfrac{56!}{\left( 50-r \right)!}}{\dfrac{54!}{\left( 51-r \right)!}}=\dfrac{30800}{1}
⇒56!(50−r)!×(51−r)!54!=30800\Rightarrow \dfrac{56!}{\left( 50-r \right)!}\times \dfrac{\left( 51-r \right)!}{54!}=30800 - (2)
We can write 56!=56×55×54!56!=56\times 55\times 54!
Similarly we can write, (51−r)!=(51−r)(50−r)!\left( 51-r \right)!=\left( 51-r \right)\left( 50-r \right)!
Now put these values in (1) and simplify it

& \dfrac{56\times 55\times 54!}{\left( 50-r \right)!}\times \dfrac{\left( 51-r \right)\left( 50-r \right)!}{54!}=30800 \\\ & 56\times 55\times \left( 51-r \right)=30800 \\\ & \Rightarrow 51-r=\dfrac{30800}{56\times 55}=10 \\\ & \therefore 51-r=10 \\\ & r=51-10=41 \\\ \end{aligned}$$ Thus we got the required value of r = 41. Note: Don’t confuse the formula of permutation with the formula of combination. In combination, the order does not matter. Thus the number of combinations will have the formula, n things taken r at a time: $${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$$.We can also verify the answer by substituting value r=41 in the expression $${}^{56}{{P}_{r+6}}:{}^{54}{{P}_{r+3}}=30800:1$$ and check whether L.H.S=R.H.S or not.