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Question: A person buys twelve packets of VANISH detergent. Each packet contains one coupon which bears one of...

A person buys twelve packets of VANISH detergent. Each packet contains one coupon which bears one of the letters of the word VANISH. If he shows all the letters of the VANISH, he gets one free packet(without a coupon). If he gets exactly one free packet then the number of different possible combinations of the coupons is

A) 11C5   {}^{11}{C_5}\,\,\,

B) 461461

C) 17C5{}^{17}{C_5}

D) 462462

Explanation

Solution

The objective is to find the chances to get the detergent packets consists the letters V,A,N,I,S,HV,A,N,I,S,H out of twelve detergent packets because if he shows all the six letters he get a free packet. To get only one free packet, the word “VANISH” should be formed only once. We assign this as V\,A\,N\,I\,S\,H\,\\_\,\\_\,\\_\,\\_\,\\_\,\\_ . On the remaining blanks we have to do combinations.

Complete step-by-step answer:

It is given that a person buys twelve packets of VANISH detergent . Each packet contains one coupon which bears one of the letters of the word VANISH. He gets a detergent packet if he shows all the letters of the given word.

Now , we have to solve that if he gets one free packet then the number of different possible combinations of coupons.

We have 1212 packets and each packet has a coupon which means we get 1212 coupons.

On each coupon, we get V,A,N,I,S,HV,A,N,I,S,H. But to get one free packet we get V,A,N,I,S,HV,A, N, I, S, H words completely once in 6 packets. And in the remaining 6 packets we can get any coupons except the exact letters V,A,N,I,S,HV,A,N,I,S,H because if we get these letters again, the person will get two extra packets freely which is not asked in the question.

V\,A\,N\,I\,S\,H\,\\_\,\\_\,\\_\,\\_\,\\_\,\\_

Each of the twelve blanks have six options of letters V,A,N,I,S,H

Let x1,x2,x3,x4,x5,x6{x_1},{x_2},{x_3},{x_4},{x_5},{x_6} denotes number of times the individual letters V,A,N,I,S,H appear on the coupons.

We write it as x1+x2+x3+x4+x5+x6=12→(1){x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} = 12 \to \left( 1 \right)

Where , 1⩽x1,x2,x3,x4,x5,x6⩽121 \leqslant {x_1},{x_2},{x_3},{x_4},{x_5},{x_6} \leqslant 12

The required number of combinations of coupons is equivalent to the number of positive integral solutions of equation 1

Now x1+x2+x3+x4+x5+x6=n{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} = n

We use the formula n−1Cr−1{}^{n - 1}{C_{r - 1}}

Apply this formula in equation (1) , we get

⇒\Rightarrow 12−1C6−1=11C5=462{}^{12 - 1}{C_{6 -1 }} = {}^{11}{C_5}=462

Now we are going to subtract 11 from the above result as V,A,N,I,S,HV,A,N,I,S,H combination of letters is also a possibility to occur again but only once it is required.

=462−1=462-1

=461=461

If the person gets exactly one free packet then the number of different possible combinations of the coupons is 461461. Thus , option (B) is correct.

Note: We use the formula n−1Cr(−1){}^{n - 1}{C_{r (- 1)}} . This formula is applied when we have to find the number of non negative integral solutions of the equation. This equation is represented as x1+x2+x3+. . .+xr=n{x_1} + {x_2} + {x_3} + .\,.\,. + {x_r} = n where r is the number of variables and n is the sum of all these variables.